text:
How to name an unsaturated spiro compound with many substituents?

I've been trying to name this compound, but I just can't, I've gotten a brilliant answer already, it's(8E,11Z,15Z)-4-amino-11-carbamoyl-3-chloro-2-(chlorocarbonyl)-9-cyano-5-formyl-4-hydroxy-13-methoxy-12-(methoxycarbonyl)-10-oxospiro[6.9]hexadeca-2,8,11,15-tetraene-8,14-dicarboxylic acid. I just want to know if there's any contrary opinion from the answer.

What have you been trying? Name the parent structure as spiro compound, identify the principal characteristic group, and add the other substituents as usual. Where are you having problems?

@Loong: I have been trying to identify the principal characteristic group, and how to number the other substitutents

You should edit your question and add your approach and preliminary result(s). Otherwise it might get closed as 'homework-like question without own efforts'

Hmm, you're sure all double bonds in larger ring are cis? Such isomer could be less stable.

The parent structure is a spiro compound, which has a fixed numbering starting in the smaller ring.
The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-24.2.1 Monospiro alicyclic ring systems
Mono spiro parent hydrides consisting of two saturated cycloalkane rings are named by placing the nondetachable prefix ‘spiro’ before the name of the unbranched acyclic hydrocarbon with the same total number of skeletal atoms. The number of skeletal atoms linked to the spiro atom in each ring is indicated by arabic numbers separated by a full stop, cited in ascending order and enclosed in square brackets; this descriptor (called in these recommendations the ‘von Baeyer spiro descriptor’) is placed between the spiro prefix and the name of the acyclic alkane. Numbering starts in the smaller ring, if one is smaller, at a ring atom next to the spiro atom and proceeds first around that ring, then through the spiro atom and around the second ring.

In this case, the name of the parent spiro hydrocarbon ring structure is spiro[6.9]hexadecane.
The fixed numbering of the spiro hydrocarbon ring is not modified by the introduction of the double bonds and functional groups. However, low locants must be attributed if there is a choice.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(a) fixed numbering in chains, rings, or ring systems, i.e., when the numbering of a system is fixed, (…), this numbering must be used, both in PINs and in general nomenclature;
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Thus, low locants are first assigned to the principal characteristic group, which is carboxylic acid. This leads to the name spiro[6.9]hexadecane-8,14-dicarboxylic acid.
Next, locants are assigned to the double bonds according to Rule (e), which yields (8E,11Z,15Z)-spiro[6.9]hexadeca-2,8,11,15-tetraene-8,14-dicarboxylic acid.
Finally, locants are assigned to all other substituents. Therefore, the correct name for the complete compound is (8E,11Z,15Z)-4-amino-11-carbamoyl-3-chloro-2-(chlorocarbonyl)-9-cyano-5-formyl-4-hydroxy-13-methoxy-12-(methoxycarbonyl)-10-oxospiro[6.9]hexadeca-2,8,11,15-tetraene-8,14-dicarboxylic acid.


text:
Are Cubic Equation of states applicable to Liquids?
I was just going through a graph of isotherm given by cubic equation of state and noticed that isotherms passes trough the subcooled liquid region. Are these equation applicable to liquids?

It's applicable, but it may not accurately describe the behavior of any real liquid.

Cubic equations of state are summarized here.  They can, indeed, describe both the liquid and gaseous state.
Under subcritical conditions, a given temperature and pressure will give three roots.  The largest is the vapor volume, the smallest is the liquid volume and the middle one is an unstable state that must exist along with the vapor and liquid roots.  By using appropriate integrated forms we can compare the free energies associated with these roots and thus identify which is more stable.  Or, we can identify when the free energies match and thus predict vapor-liquid equilibrium.  The catch?  The equations are of only limited accuracy, especially for liquids.  Some are better than others; for more details the reader can start with the Wikipedia reference given above (https://en.m.wikipedia.org/wiki/Equation_of_state).

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text:
Normalized standard calibration curve
I have a dynamic assay whose absorbance changes with time. I recorded it’s spectrum initially after preparing it. Then I added my analyte and mixed the contents. Then I waited for 2 min and recorded the second absorbance soectrum. I generated a normalized standard calibration curve by dividing the absorbance maximas (captured 2 min post analyte addition) at different analyte concentrations (1M, 2M, 3M...)  by an average value of control (initial absorbance maxima before analyte addition). Is this approach correct? My ordinate is normalized absorbance and abscissa is concentration of analyte.

I'm not sure that I understand your method entirely. If your measuring absorbance at a single wavelength, then typically I'd say that the absorbance would be measured on a blank (ie solution without analyte) and subtracted from the various solutions with analyte. If you taking a whole spectrum then you might be able to get background from each spectra, but typically UV-Vis peaks are so broad that this might not be possible. Also not sure if you have overlapping peaks. So all in all a lot more detail is needed to understand what you are doing.

It depends what you want to know and to do afterwards. Do you want to determine the order of the reaction and the rate constant ? Do you just want to check whether the reaction stays within a certain degree of advancement ? Why do you need a calibration curve ? Is the absorbance not sufficient ? Why do you measure the whole spectrum ? It should be much quicker to just measure the absorbance at the summit of the peaks. Have you tried to measure the absorbance after  more than 2 minutes ? If not, why ?

I'm not sure whether calibration for a dynamic assay is any different from one that is stable. But let's think of the situation like this. What you have done is no different from what an operator would do to obtain a calibration curve with an internal standard (IS) correction (by plotting concentration against the ratio of the intensities of analyte/IS) except you have used the absolute intensity of your background signal (no analyte) instead of an internal standard for correction. This is not correct by any standard since the background is a bad choice for correction (the background is NOT stable, and has a terrible RSD due to low counts). What you should do is to plot your calibration curve without normalising it to the background.

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text:
Why is a double bond given priority over substituents during IUPAC nomenclature?

My book names this compound as 6-ethyl-5-methylcyclohex-2-en-1-one.
I have learnt that after numbering the principal functional group which is ketone in this case, we have to treat all other groups as substituents and therefore I think the name should be 2-ethyl-3-methylcyclohex-5-en-1-one.
Why should the double bond be given priority?

Not an answer, but can confirm that several sources independently suggest that 6-ethyl-5-methylcyclohex-2-en-1-one is the correct name. Note also that the italics/formatting used in your post is superfluous!

I do not see any reason why I should give priority to the double bond here.

The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

You have correctly identified the principal characteristic group and assigned the lowest locant to the ketone in accordance with Rule (c).
Next, a low locant is assgined to the double bond according to Rule (e). Therefore, the name of the structure without further substituents is cyclohex-2-en-1-one rather than cyclohex-5-en-1-one since the locant 2 for the double bond is lower than 5.
Thus, the correct name is 6-ethyl-5-methylcyclohex-2-en-1-one.

This helps. So While naming I must:  Assign the lowest locant to the principal functional group and then look for double/multiple bonds and give them the next priority. Right?

@ParthChauhan Yes, for most simple organic compounds (in particular, compounds without added indicated hydrogen) this is correct.


text:
Reduction of 2-cyclohexenone to cyclohexanol
Is $\ce{NaBH4}$ a suitable reducing agent for the conversion of 2-cyclohexenone to cyclohexanol?
In Clayden (page 506, second edition), I have found the reduction of cyclopentenone to cyclopentanol using $\ce{NaBH4}$ in $\ce{Et-OH}$. (99% yield!). First NaBH4 performs 1,4 addition and then 1,2 addition to give the product.
However, I am unable to find a reference for reduction of cyclohexenone to cyclohexanol.
I found this in Smith:

I don't understand why it can't reduce the double bond through Michael addition in case of cyclohex-2-enone as well. What is the reason behind it?

Whether NaBH4 reduces the C=C double bond seems to be substrate dependent (and more often than not, it actually doesn't reduce it). And I am not sure if there is any pattern to it.

According to Johnson, M. R.; Rickborn B. Sodium borohydride reduction of conjugated aldehydes and ketones J. Org. Chem., 1970, 35, 1041.
DOI: 10.1021/jo00829a039:
2-Cyclohexenone can certainly be reduced fully. It has very similar reactivity to cyclopentenone. In this report, mixtures of products were observed. The experimental indicates the researchers used 0.5 eq of $\ce{NaBH4}$ per carbonyl.

Furthermore, the use of Luche conditions can again help you favor only 1,2 reduction:

I am just curious. Could you elaborate on the principle of Luche reduction? What effect does the cerium (III) ion have on the conjugated enone?

@TanYongBoon this site has a good explanation: https://www.organic-chemistry.org/namedreactions/luche-reduction.shtm


text:
Why is perchloric acid stronger than perbromic acid?
Since chlorine has a larger electronegativity than bromine, why is perchloric acid a strong acid, while perbromic acid is not?  We clearly see that electronegativity affects an acid's strength with hydrofluoric acid being much weaker than hydrochloric acid, due to fluorine's high electronegativity.  Wouldn't chlorine's higher electronegativity make it harder for perchloric acid to dissociate than perbromic acid like it does for hydrofluoric acid?


text:
Which formula can be used to calculate the exact hydronium concentration present in sodium hydrogen carbonate solution?
Which formula  can be used to calculate the exact  hydronium concentration present in  sodium hydrogen carbonate solution ?

$$\ce{[H3O+]}=\sqrt{K_\mathrm{a1}K_\mathrm{a2}}$$
$$[\ce{H3O+}]=\sqrt{\frac{K_\mathrm{a1}K_\mathrm{w}}{[\ce{NaHCO3}]}}$$
$$[\ce{H3O+}] = \sqrt{\frac{K_\mathrm{a1}K_\mathrm{a2}\times[\ce{NaHCO3}]+K_\mathrm{a1}K_\mathrm{w}}{[\ce{NaHCO3}]+K_\mathrm{a1}}}$$

I  tried to order the series  of reactions once sodium hydrogen carbonate  is dissociated according  the following sequence :

Taking into consideration  equation two and equation four, and  taking the assumption: Degree of ionization=degree of hydrolysis, an approximate calculation hydronium concentration can be done by the  formula  one.
Recognizing  equation four and equation two  , that for every one mole of carbonate- formed, one mole of hydronium was formed, and  for every mole of carbonic acid formed, one mole of hydronium  was used up and when comparing K value between both of these equation, I can ignore the contribution of equation four. And because of the low concentration of carbonic acid  in equation three and low carbonate ion  concentration in equation five , I can ignore the contribution of  both equation. So I take into my consideration equation two to derive formula two.
I tried to  derive formula three but  I don't know the assumptions beyond this formula and how to derive it.
I need to know how  to calculate concentrations  of  all species are present in  sodium hydrogen carbonate solution ?
I appreciate any help.

Well none of the formulas calculate the exact pH. One will calculate an  approximate pH, assuming $\ce{[HCO3^-] > 10^{-6}}$ so autoionization of water is irrelevant. Also assume that $\ce{[CO3^{2-}] \approx  [H2CO3]}$

@MaxW 2According  to the two assumption  mentioned  above in the comment : pH is  independent  of salt concentration ,so we choose formula one , but how to calculate pH without the second assumption?

It is a mess. See Section 9 starting on page 42 of Acid-base Equilibria and Calculations. In a nut shell the complete solution considers the autodissociation of water, and the concentrations of species CO2(g), H2CO3, HCO3-, and CO3(-2). Take a look at figure 16.

Note that by assuming $\ce{[CO3^{2-}] = [H2CO3] }$ the pH is nailed to 8.31. The more realistic assumption is that $\ce{[HCO3-] \gg [CO3^{2-}] + [H2CO3] }$ which means that pH will vary with changes in initial concentration of $\ce{[HCO3-]}$. (This still neglects autodissociation of water.)

@MaxW Thank you,now I understand the  assumptions which relate formula one with formula three ,  but I don’t know  how to relate (reconcile) formula two to  formula three?

@MaxWI use formula  two in teaching , is it accepted?  .

@MaxWBecause the $K_{\mathrm b}$ of $\ce{HCO3-}$ is larger than $K_{\mathrm a}$ of $\ce{HCO3-}$ , I neglect reaction four and depend on equilibrium expression of reaction two to derive formula two .

@AdnanALAmleh - I'd be real careful when mixing Ka and Kb equations for a problem like this. Personally I like to use either the two Ka formulas or the two Kb formulas depending on if the solution is going to be acidic or basic and then the explicitly use the Kw expression if needed. The chemical equations in 2-4 just contain three different formulas not four. Three because you can get $K_w$ from a Ka/Kb pair.

@MaxW This question keeps coming up on the front page stating that you did some activity ("modified") on the page (for example, 8 hours ago right now).  However, I can't see any activity here. Are you editing a deleted answer?

@GaurangTandon - Yes. Didn't remember that editing a "deleted" answer would cause that. I guess I really need to get an off-line editing to work. I just can't visualize the formula and equations markup. In the meantime, I'll edit, cut and past into a text file on my computer and "cancel" the edit. I think that will solve the problem.

@MaxW If it's a problem with the typesetting in MathJax, I can help you with that :) Just post your answer and I'll edit it. Otherwise, feel free to continue saving the edits to the deleted answer anyway. The front page bumps are not a big deal. Cheers!

@GaurangTandon - Thanks for the offer. I know how to do the edits (mostly...) but the markups get so long that I just can't see a mistake. Also I find it the text parts hard to read with all the markup intertwined.

Wording of the Question...
The question asks: Which formula can be used to calculate the exact hydronium concentration present in sodium hydrogen carbonate solution?
What does "exact" mean?
The word exact gives me pause. In general there isn't usually an exact answer, but rather only an answer that is within the required experimental error. The values of $K_\mathrm{a1}$, $K_\mathrm{b1}$, and $K_\mathrm{b2}$ are given to three significant figures. The value for $K_\mathrm{a2}$ is only given to two significant figures, but
$$K_\mathrm{a2} = \dfrac{K_\mathrm{w}}{K_\mathrm{b2}} = \dfrac{1.00\times10^{-14}}{2.13\times10^{-4}} = 4.69483\times10^{-11}\tag{SF1}$$
So it seems reasonable to assume that $K_\mathrm{a2} = 4.70\times10^{-11}$ and that exact means to within 1 part per thousand (ppt).
Since the pH of pure water is taken to be 7.000, this means that for any pH between 4.000 and 10.000 the autoionization of water will have to be taken into account to get results within 1 ppt. (Remember that the characteristic of the pH doesn't count in significant figures, only the mantissa does.)
Realistically that is terribly precise. At 25 $^\circ \mathrm{C}$ $K_w \approx 13.995$ and the change is about 0.033 per $^\circ \mathrm{C}$. So a 1 ppt change is only a 0.03 $^\circ \mathrm{C}$ change in temperature. For such calculations I'd probably expect the typical error would be on the order of ± 5% of reality. But the problem poses 1 ppt precision and so that is how the calculations "should" be done. If the problem poser wanted less precision them the equilibrium constants should have been specified with fewer significant figures.
Setup
Assumptions

Ignoring relationship between $\ce{CO2}$ gas phase and dissolved $\ce{CO2}$ in the liquid phase. Mass balance would require relative volumes of gas and liquid phases.
Ignoring relationship between dissolved $\ce{CO2}$ and carbonic acid. Accept  that $K_{\mathrm{a1}}$ is the "combined" equilibrium constant which takes that relationship into account.
The equilibriums will be calculated using concentrations instead of the more appropriate activities. So above 1 millimolar concentration of sodium bicarbonate the calculated results are suspect.

Starting with the basic relationships.
$$K_{\mathrm{a1}} = \dfrac{\ce{[H+][HCO3^-]}}{\ce{[H2CO3]}}\tag{S1}$$
$$K_{\mathrm{a2}} = \dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^-]}}\tag{S2}$$
$$K_\mathrm{w} = \ce{[H^-][OH^-]}\tag{S3}$$
Letting $\mathrm{[C_T]}$ be the total concentration of all the various carbonate species (which we know from is nominal molarity of the sodium bicarbonate).
$$\mathrm{[C_T]} = \ce{[H2CO3] + [HCO3^-] + [CO3^{2-}]}\tag{S4}$$
But this leaves us with five unknowns ($\ce{[H+], [OH^-], [H2CO3], [HCO3^-]}$ and $\ce{[CO3^{2-}]}$) but only four equations (S1), (S2), (S3) and (S4). So another equation is needed relating the five unknowns to completely numerically solve the system.
Fraction of Species
The fraction of each species can be calculated for a particular pH value using equations (F1), (F2), and (F3). Basically you substitute values into (S4) as a function of $\ce{[H+]}$ and whatever species ($\ce{[H2CO3], [HCO3^-], [CO3^{2-}]}$) that you're interest in. The equations for the fractions of the species are exact.
$$\dfrac{\ce{[H_2CO_3]}}{\ce{[C_T]}} = \dfrac{\ce{[H^+]^2}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F1}$$
$$\dfrac{\ce{[HCO3^-]}}{\ce{[C_T]}} = \dfrac{\ce{K_\mathrm{a1}[H^+]}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F2}$$
$$\dfrac{\ce{[CO3^{2-}]}}{\ce{[C_T]}} = \dfrac{\ce{K_\mathrm{a1}K_\mathrm{a2}}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F3}$$
Think of dissolving $\ce{H2CO3}$ ($\ce{C_T}$) and then adding between 0 and 2 equivalents of $\ce{NaOH}$. The equations relate the three variables pH, $\ce{[C_T]}$ and the $x$ equivalents of $\ce{NaOH}$. Knowing any two fixes the third.
Another interesting fact is found by differentiating equation (F2) for $\ce{[H^+]}$ which shows that the maximum of $\ce{[HCO3^-]/[C_T]}$ (or any diprotic acid...) occurs when
$$\ce{[H^+] = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}}\tag{F4}$$
It is the pH as which the bicarbonate solution has its maximum buffering capacity. Also at that pH $\ce{[H2CO3] = [CO3^{2-}]}$.
For the bicarbonate system the maximum concentration of $\ce{HCO3^-}$ is 97.99% which occurs at a pH of 8.340. At that pH $\ce{[H2CO3] = [CO3^{2-}] = 1.01\%}$. So for an accuracy of 1 ppt the hydrolysis of water and the multiple species cannot be neglected.
If the pH is above 7.648, then the concentration of the bicarbonate species will be at least 95%.
\begin{array}{|c|c|c|c|c|}\hline
\ce{[H+]} & \mathrm{pH} & \ce{F(H2CO3)} & \ce{F(HCO3^-)}&\ce{F(CO3^{2-})}\\ \hline
1.00E+00 & 0.000E+00 & 1.000E+00 & 4.450E-07 & 2.091E-17 \\ \hline
8.00E-01 & 9.691E-02 & 1.000E+00 & 5.562E-07 & 3.268E-17 \\ \hline
6.00E-01 & 2.218E-01 & 1.000E+00 & 7.417E-07 & 5.810E-17 \\ \hline
4.00E-01 & 3.979E-01 & 1.000E+00 & 1.112E-06 & 1.307E-16 \\ \hline
2.00E-01 & 6.990E-01 & 1.000E+00 & 2.225E-06 & 5.229E-16 \\ \hline
1.00E-01 & 1.000E+00 & 1.000E+00 & 4.450E-06 & 2.091E-15 \\ \hline
8.00E-02 & 1.097E+00 & 1.000E+00 & 5.562E-06 & 3.268E-15 \\ \hline
6.00E-02 & 1.222E+00 & 1.000E+00 & 7.417E-06 & 5.810E-15 \\ \hline
4.00E-02 & 1.398E+00 & 1.000E+00 & 1.112E-05 & 1.307E-14 \\ \hline
2.00E-02 & 1.699E+00 & 1.000E+00 & 2.225E-05 & 5.229E-14 \\ \hline
1.00E-02 & 2.000E+00 & 1.000E+00 & 4.450E-05 & 2.091E-13 \\ \hline
8.00E-03 & 2.097E+00 & 9.999E-01 & 5.562E-05 & 3.268E-13 \\ \hline
6.00E-03 & 2.222E+00 & 9.999E-01 & 7.416E-05 & 5.809E-13 \\ \hline
4.00E-03 & 2.398E+00 & 9.999E-01 & 1.112E-04 & 1.307E-12 \\ \hline
2.00E-03 & 2.699E+00 & 9.998E-01 & 2.225E-04 & 5.228E-12 \\ \hline
1.00E-03 & 3.000E+00 & 9.996E-01 & 4.448E-04 & 2.091E-11 \\ \hline
8.00E-04 & 3.097E+00 & 9.994E-01 & 5.559E-04 & 3.266E-11 \\ \hline
6.00E-04 & 3.222E+00 & 9.993E-01 & 7.411E-04 & 5.805E-11 \\ \hline
4.00E-04 & 3.398E+00 & 9.989E-01 & 1.111E-03 & 1.306E-10 \\ \hline
2.00E-04 & 3.699E+00 & 9.978E-01 & 2.220E-03 & 5.217E-10 \\ \hline
1.00E-04 & 4.000E+00 & 9.956E-01 & 4.430E-03 & 2.082E-09 \\ \hline
8.00E-05 & 4.097E+00 & 9.945E-01 & 5.532E-03 & 3.250E-09 \\ \hline
6.00E-05 & 4.222E+00 & 9.926E-01 & 7.362E-03 & 5.767E-09 \\ \hline
4.00E-05 & 4.398E+00 & 9.890E-01 & 1.100E-02 & 1.293E-08 \\ \hline
2.00E-05 & 4.699E+00 & 9.782E-01 & 2.177E-02 & 5.115E-08 \\ \hline
1.00E-05 & 5.000E+00 & 9.574E-01 & 4.260E-02 & 2.002E-07 \\ \hline
8.00E-06 & 5.097E+00 & 9.473E-01 & 5.269E-02 & 3.096E-07 \\ \hline
6.00E-06 & 5.222E+00 & 9.310E-01 & 6.905E-02 & 5.409E-07 \\ \hline
4.00E-06 & 5.398E+00 & 8.999E-01 & 1.001E-01 & 1.176E-06 \\ \hline
2.00E-06 & 5.699E+00 & 8.180E-01 & 1.820E-01 & 4.277E-06 \\ \hline
1.00E-06 & 6.000E+00 & 6.920E-01 & 3.080E-01 & 1.447E-05 \\ \hline
8.00E-07 & 6.097E+00 & 6.426E-01 & 3.574E-01 & 2.100E-05 \\ \hline
6.00E-07 & 6.222E+00 & 5.741E-01 & 4.258E-01 & 3.336E-05 \\ \hline
4.00E-07 & 6.398E+00 & 4.733E-01 & 5.266E-01 & 6.187E-05 \\ \hline
2.00E-07 & 6.699E+00 & 3.100E-01 & 6.898E-01 & 1.621E-04 \\ \hline
1.00E-07 & 7.000E+00 & 1.834E-01 & 8.162E-01 & 3.836E-04 \\ \hline
8.00E-08 & 7.097E+00 & 1.523E-01 & 8.472E-01 & 4.977E-04 \\ \hline
6.00E-08 & 7.222E+00 & 1.187E-01 & 8.806E-01 & 6.898E-04 \\ \hline
4.00E-08 & 7.398E+00 & 8.239E-02 & 9.165E-01 & 1.077E-03 \\ \hline
2.25E-08 & 7.648E+00 & 4.803E-02 & 9.500E-01 & 1.984E-03 \\ \hline
2.00E-08 & 7.699E+00 & 4.291E-02 & 9.548E-01 & 2.244E-03 \\ \hline
1.00E-08 & 8.000E+00 & 2.188E-02 & 9.735E-01 & 4.576E-03 \\ \hline
8.00E-09 & 8.097E+00 & 1.756E-02 & 9.767E-01 & 5.738E-03 \\ \hline
6.00E-09 & 8.222E+00 & 1.320E-02 & 9.791E-01 & 7.670E-03 \\ \hline
4.57E-09 & 8.340E+00 & 1.006E-02 & 9.799E-01 & 1.008E-02 \\ \hline
4.00E-09 & 8.398E+00 & 8.806E-03 & 9.797E-01 & 1.151E-02 \\ \hline
2.00E-09 & 8.699E+00 & 4.372E-03 & 9.728E-01 & 2.286E-02 \\ \hline
1.00E-09 & 9.000E+00 & 2.142E-03 & 9.531E-01 & 4.479E-02 \\ \hline
8.00E-10 & 9.097E+00 & 1.695E-03 & 9.429E-01 & 5.540E-02 \\ \hline
6.00E-10 & 9.222E+00 & 1.249E-03 & 9.262E-01 & 7.255E-02 \\ \hline
4.00E-10 & 9.398E+00 & 8.037E-04 & 8.941E-01 & 1.051E-01 \\ \hline
2.00E-10 & 9.699E+00 & 3.638E-04 & 8.094E-01 & 1.902E-01 \\ \hline
1.00E-10 & 1.000E+01 & 1.528E-04 & 6.802E-01 & 3.197E-01 \\ \hline
8.00E-11 & 1.010E+01 & 1.132E-04 & 6.298E-01 & 3.700E-01 \\ \hline
6.00E-11 & 1.022E+01 & 7.560E-05 & 5.607E-01 & 4.392E-01 \\ \hline
4.00E-11 & 1.040E+01 & 4.133E-05 & 4.598E-01 & 5.402E-01 \\ \hline
2.00E-11 & 1.070E+01 & 1.342E-05 & 2.985E-01 & 7.015E-01 \\ \hline
1.00E-11 & 1.100E+01 & 3.942E-06 & 1.754E-01 & 8.246E-01 \\ \hline
\end{array}
The fourth equation
The fourth equation comes from the charge balance in the solution. The solution must be electrically neutral so the total charge on cations and anions must be equal.
$$\ce{[Na^+] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C1}$$
But in a solution of $\ce{[NaHCO3]}$ the $\ce{[Na^+] = [C_T]}$ so:
$$\ce{[C_T] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C2}$$
Which yields and exact solution as a quartic equation. But rather than solve only that particular quartic equation, let's consider the more general case. Assume we start with $\ce{H2CO3}$ in water and add amounts of $\ce{NaOH} according to the equation:
$$\ce{H2CO3 + xNaOH -> Na_xH_{2-x}CO3 + xH2O}\tag{C3}$$
where $x$ is a real number bounded by $ 0 \le x \le 2$. The special cases of $x = 0,1,2$ correspond to adding $\ce{H2CO3, NaHCO3}$ or $\ce{Na2CO3}$ to pure water. Then it is intuitively obvious to the casual observer that (a la analytical chemist Dr. Peter Carr, University of Ga.) the general equation is:
$$\begin{multline}
  [\ce{H+}]^4
  + (K_\mathrm{a1} + xC_T)[\ce{H+}]^3
  + (K_\mathrm{a1} K_\mathrm{a2} + (x - 1) C_T K_\mathrm{a1}
    - K_\mathrm{w})[\ce{H+}]^2\\
  + K_\mathrm{a1} ((x - 2)C_T K_\mathrm{a2} - K_\mathrm{w})[\ce{H+}]
  - K_\mathrm{a1} K_\mathrm{a2} K_\mathrm{w}
  \tag{C4}
\end{multline}$$
It is not hard for a computer to solve the quartic equation numerically, but to do so by hand is extremely tedious. Hence books never(?) required such a solution in problem sets.
DERIVATION OF EQUATION 1 $\ce{[H^+]} = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}$
$$K_\mathrm{a1}K_\mathrm{a2} = \dfrac{\ce{[H+][HCO3^-]}}{\ce{[H2CO3]}}\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^-]}} = \dfrac{\ce{[H+]^2[CO3^{2-}]}}{\ce{[H2CO3]}}\tag{EQ1-1}$$
rearrange (EQ1-1)
$$\ce{[H+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[H2CO3]}}{\ce{[CO3^{2-}]}}}\tag{EQ1-2}$$
assume $\ce{[H2CO3] = [CO3^{2-}]}$
Then
$$\ce{[H+]} = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}\tag{EQ1-3}$$
DERIVATION OF EQUATION 2 $\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T]}}}$
Sodium is a spectator ion. Thus it is be possible to model the system as a monoprotic Brønsted–Lowry acid which has $\ce{H^+}$ as the acid and $\ce{HCO3^-}$ as the conjugate base.
This derivation has two problems. First it ignores the autoionization of water. Second the equation models the system as a monoprotic base, but the carbonate system has three species - carbonic acid, bicarbonate, and carbonate, and all three are likely to be present in significant quantities in alkaline solutions.
Start with the reaction of the conjugate base
$$\ce{HCO3^- + H2O <=> H2CO3 + OH^-}\tag{EQ2-1}$$
and the equilibrium expression
$$\ce{K_\mathrm{b1}} = \dfrac{\ce{[H2CO3][OH^-]}}{\ce{[HCO3^-]}}\tag{EQ2-2}$$
Assume that little of $\ce{HCO3^-}$ accepts a proton, thus $\ce{[HCO3^-] = [C_T]}$ and $\ce{[H2CO3] = [OH^-]}$. Now substituting:
$$\ce{K_\mathrm{b1}} = \dfrac{\ce{[OH^-]^2}}{\ce{[C_T]}}\tag{EQ2-3}$$
but $\ce{K_\mathrm{b1}} = K_w / K_{a1}$ and $\ce{[OH^-] = K_w / [H^+]}$. Now making more substitutions
$$\dfrac{K_\mathrm{w}}{K_\mathrm{a1}} = \dfrac{(\frac{K_\mathrm{w}}{\ce{[H^+]}})^2}{[C_T]}\tag{EQ2-4}$$
rearranging and solving for $\ce{[H^+]}$ gives the desired result.
$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T]}}}\tag{EQ2-5}$$
The desired result could have been obtained more expeditiously by starting with equation (2) and letting $\ce{[H2CO3] = [OH^-] = K_w/[H^+]}$, but that route doesn't seem as clear chemically and seems more like math magic.
DERIVATION OF EQUATION 3 $\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}C_T + K_{a1}K_w}{C_T+K_{a1}}}$
Starting with equation (C2) for the charge balance of a solution of $\ce{NaHCO3}$ where $\ce{[C_T] = [Na^+]}$
$$\ce{[C_T] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C2}$$
subtract equation (S4) which is for mass balance of all the carbonate species
$$\mathrm{[C_T]} = \ce{[H2CO3] + [HCO3^-] + [CO3^{2-}]}\tag{S4}$$
yielding
$$\ce{[H^+] = [CO3^{2-}] - [H2CO3] + [OH^-]}\tag{EQ3-1}$$
rearranging
$$\ce{[OH^-] = [H2CO3] - [CO3^{2-}] + [H^+]}\tag{EQ3-2}$$
substituting $K_\mathrm{w} / \ce{[H^+] = [OH^-]}$ and rearrange
$$\dfrac{K_\mathrm{w}}{\ce{[H^+]}} \ce{+ [CO3^{2-}] = [H2CO3] + [H^+]}\tag{EQ3-3}$$
Solve equation (S1) for $\ce{[H2CO3]}$ and solve equation (S2) for $\ce{[CO3^{2-}]}$, and then substitute those results into equation (EQ3-3)
$$\dfrac{K_\mathrm{w}}{\ce{[H^+]}} + \dfrac{K_\mathrm{a2}\ce{[HCO3^-]}}{\ce{[H^+]}} = \dfrac{\ce{[H^+][HCO3^-]}}{K_\mathrm{a1}} + \ce{[H^+]}
\tag{EQ3-4}$$
Multiplying both sides by $\ce{[H^+]}$, then $K_\mathrm{a1}$, and finally collecting terms
$$K_\mathrm{a1}K_\mathrm{w} + K_\mathrm{a1}K_\mathrm{a2}\ce{[HCO3^-]} = \ce{[H^+]^2([HCO3^-] + K_\mathrm{a1})}
\tag{EQ3-5}$$
rearranging and solving for $\ce{[H^+]}$ yields the exact equation for $\ce{[H^+]}$ in terms of the particular species $\ce{[HCO3^-]}$ when $\ce{NaHCO3}$ is dissolved in pure water.
$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[HCO3^-]} + \mathrm{K_{a1}K_w}}{\ce{[HCO3^-] + \mathrm{K_{a1}}}}}\tag{EQ3-6}$$
As indicated previously the above equation is exact in terms of the relationship between $\ce{[H^+]}$ and the carbonate species $\ce{HCO3^-}$. Over the whole pH range $\ce{[C_T] \ne [HCO3^-]}$, however for the range $ 7.648 \lt \mathrm{pH} \lt 9.00$ the species $\ce{HCO3^-}$ is at least 95% of $\ce{[C_T]}$. So we can "reasonably" substitute $\ce{[C_T] = [NaHCO3]}$ and get the desired equation even though the equation is not exact.
$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[C_T]} + K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T] + K_\mathrm{a1}}}}\tag{EQ3-7}$$
!! CHECKS !!
Why aren't students taught to do checks any more!?!
The column $\ce{[C_T]}$ is the nominal molarity of the sodium bicarbonate solution. It is also the molarity of the spectator cation $\ce{[Na^+]}$.
The column $\Sigma \space \text{cations}$ should equal the column $\Sigma \space \text{anions}$.
The column % Diff is a measure of the inconsistency of the calculations of the concentration of anions and cations and is calculated by
$$\text{% Diff} = 200*\dfrac{\Sigma\text{ cations} - \Sigma\text{ anions}}{\Sigma\text{ cations} + \Sigma\text{ anions}}$$
Again, these calculations use concentrations instead of the more appropriate activities. Concentrations are probably fine for these solutions below 1 millimolar. Above that and activities should really be used.
EQUATION 1 $\ce{[H^+]} = \sqrt{K_{a1}K_{a2}}$
\begin{array}{|c|c|c|c|c|c|c|}\hline
\ce{[C_T]} & calc \space [H^+] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff\\ \hline
1.00E-01 & 4.5733E-09 &  8.340 &  1.0000E-01&  2.1866E-06 &9.7986E-02 &2.0140E-03 &1.0000E-01 & 0.0 \\ \hline
1.00E-02 &4.5733E-09 &  8.340 &1.0000E-02 &2.1866E-06 &9.7986E-03 &2.0140E-04 &1.0002E-02 & 0.0 \\ \hline
1.00E-03 &4.5733E-09 &  8.340 &1.0000E-03 &2.1866E-06 &9.7986E-04 &2.0140E-05 &1.0022E-03 & -0.2 \\ \hline
1.00E-04 &4.5733E-09 &  8.340 &1.0000E-04 &2.1866E-06 &9.7986E-05 &2.0140E-06 &1.0219E-04 & -2.2\\ \hline
1.00E-05 &4.5733E-09 &  8.340 &1.0005E-05 &2.1866E-06 &9.7986E-06 &2.0140E-07 &1.2187E-05 & -19.7\\ \hline
1.00E-06 &4.5733E-09 &  8.340 &1.0046E-06 &2.1866E-06 &9.7986E-07 &2.0140E-08 &3.1866E-06 & -104 \\ \hline
1.00E-07 &4.5733E-09 &  8.340 &1.0457E-07 &2.1866E-06 &9.7986E-08 &2.0140E-09 &2.2866E-06 & -183 \\ \hline
\end{array}
So assuming that the pH is constant only works for concentrated solutions (greater than  1 millimolar) for which activities should really be used...
EQUATION 2 $\ce{[H^+]} = \sqrt{\dfrac{K_{a1}K_w}{\ce{[C_T]}}}$
\begin{array}{|c|c|c|c|c|c|c|}\hline
\ce{[C_T]} & calc \space [H^+] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff\\ \hline
1.00E-01 &2.1095E-10 & 9.676 &1.0000E-01 &4.7405E-05 &8.1748E-02 &3.6427E-02 &1.1822E-01 & -16.7 \\ \hline
1.00E-02 &6.6708E-10 & 9.176 &1.0000E-02 &1.4991E-05 &9.3287E-03 &1.3145E-03 &1.0658E-02 & -6.4\\ \hline
1.00E-03 &2.1095E-09 & 8.676 &1.0000E-03 &4.7405E-06 &9.7369E-04 &4.3388E-05 &1.0218E-03 & -2.2\\ \hline
1.00E-04 &6.6708E-09 & 8.176 &1.0001E-04 &1.4991E-06 &9.7844E-05 &1.3787E-06 &1.0072E-04 & -0.7 \\ \hline
1.00E-05 &2.1095E-08 & 7.676 &1.0021E-05 &4.7405E-07 &9.5271E-06 &4.2453E-08 &1.0044E-05 & -0.2 \\ \hline
1.00E-06 &6.6708E-08 & 7.176 &1.0667E-06 &1.4991E-07 &8.6910E-07 &1.2247E-09 &1.0202E-06 & +4.5\\ \hline
1.00E-07 &2.1095E-07 & 6.676 &3.1095E-07 &4.7405E-08 &6.7830E-08 &3.0225E-11 &1.1527E-07 & +91.8 \\ \hline
\end{array}
If 1 ppt is required, then Equation 2 doesn't work. For a +/- 5% error, the equation works in the range from $1\times10^{-2}$ to $1\times10^{-6}$ molar.
EQUATION 3 $\ce{[H^+]} = \sqrt{\dfrac{K_{a1}K_{a2}C_T + K_{a1}K_w}{C_T+K_{a1}}}$
\begin{array}{|c|c|c|c|c|c|c|}\hline
\ce{[C_T]} & calc \space [H] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff \\ \hline
1.00E-01 &4.5781E-09 & 8.339 &1.0000E-01 &2.1843E-06 &9.7986E-02 &2.0119E-03 &1.0000E-01 & 0.0 \\ \hline
1.00E-02 &4.6216E-09 & 8.335 &1.0000E-02 &2.1638E-06 &9.7986E-03 &1.9930E-04 &1.0000E-02 & 0.0 \\ \hline
1.00E-03 &5.0352E-09 & 8.298 &1.0000E-03 &1.9860E-06 &9.7977E-04 &1.8291E-05 &1.0000E-03 & 0.0 \\ \hline
1.00E-04 &8.0700E-09 & 8.093 &1.0001E-04 &1.2392E-06 &9.7660E-05 &1.1376E-06 &1.0004E-04 & 0.0 \\ \hline
1.00E-05 &2.1120E-08 & 7.675 &1.0021E-05 &4.7348E-07 &9.5267E-06 &4.2400E-08 &1.0043E-05 & -0.2\\ \hline
1.00E-06 &5.5624E-08 & 7.255 &1.0556E-06 &1.7978E-07 &8.8822E-07 &1.5010E-09 &1.0695E-06 & -1.3\\ \hline
1.00E-07 &9.0382E-08 & 7.044 &1.9038E-07 &1.1064E-07 &8.3082E-08 &8.6408E-11 &1.9381E-07 & -1.8 \\ \hline
1.00E-08 &9.8897E-08 & 7.005 &1.0890E-07 &1.0111E-07 &8.1785E-09 &7.7735E-12 &1.0930E-07 & -0.4\\ \hline
\end{array}
If 1 ppt is required, then Equation 3 works for concentrations grater than  $1\times10^{-5}$ molar. For a +/- 5% error, the equation works in the range from $0.1$ to $1\times10^{-8}$ molar.
Summary of $\ce{[H^+]}$ values
the table below is a summary of the results for the three derived equations, and the exact solution using equation (C4) by Wolfram Alpha.
\begin{array}{|c|c|c|c|c|}\hline
\ce{[C_T]} & Eq(1)& Eq(2) & Eq(3) & exact\\ \hline
1.00E-01  & 4.5733E-09 & 2.1095E-10 & 4.5781E-09 & 4.5782E-9 \\ \hline
1.00E-02  & 4.5733E-09 & 6.6708E-10 & 4.6216E-09 & 4.6226E-9 \\ \hline
1.00E-03  & 4.5733E-09 & 2.1095E-09 & 5.0352E-09 & 5.0443E-9 \\ \hline
1.00E-04  & 4.5733E-09 & 6.6708E-09 & 8.0700E-09 & 8.1354E-9 \\ \hline
1.00E-05  & 4.5733E-09 & 2.1095E-08 & 2.1120E-08 & 2.1602E-8 \\ \hline
1.00E-06  & 4.5733E-09 & 6.6708E-08  & 5.5624E-08 & 5.7983E-8 \\ \hline
1.00E-07  & 4.5733E-09 & 2.1095E-07  & 9.0382E-08 & 9.1835E-8 \\ \hline
1.00E-08  & -- & -- & 9.8897E-08 & 9.9096E-8 \\ \hline
\end{array}

I am fascinated with your powerful answer which remediates my misconception about the assumptions and conditions at which each formula work.

@AdnanAL-Amleh - Added exact solution in last table. Bit of a struggle to Wolfram to work.

I appreciate  your effort and your work and I thank you

@MaxWI think the fourth algebraic term in equation ($C_4$) is as follows:$$\begin{multline}
[\ce{H+}]^4

(K_\mathrm{a1} + xC_T)[\ce{H+}]^3
(K_\mathrm{a1} K_\mathrm{a2} + (x - 1) C_T K_\mathrm{a1}

K_\mathrm{w})[\ce{H+}]^2\

K_\mathrm{a1} ((x - 2)C_T{\color{red}{K_\mathrm{a2}}} - K_\mathrm{w})[\ce{H+}]

K_\mathrm{a1} K_\mathrm{a2} K_\mathrm{w}
\tag{C4}
\end{multline}$$   correct  my comment?

@AdnanAL-Amleh - You could be right. I threw all my notes on this away a while back. I'll have to check that again.

@AdnanAL-Amleh - you're correct. equation 7  here http://www.aqion.de/site/184


text:
Formation of acid chlorides
In Sn2 nucleophilic substitution reactions,the attacking nucleophile or leaving group is eliminated. The weaker base is eliminated. Then how do acid chlorides form as OH- is a stronger base than Cl–.

Acid chlorides are normally formed from carboxylic acids by the action of reagents such as thionyl chloride (SOCl2). This is not a SN2 reaction well not at carbon.
Normally a carboxylic oxygen attacks the sulfur of the thionyl chloride displacing an chloride from the thionyl chloride. After loss of a hydrogen ion from the carboxylic acid we will have
R-C(=O)-O-S(=O)-Cl
A chloride anion then attacks the carbonyl carbon as a nucelophile, the sulfur group acts as a leaving group. It will form sulfur dioxide and another chloride anion. In terms of the reaction at the organic part of the molecule it should be classified as a nucelophilic attack on a carbonyl compound which results in the substitution of one group for another. In terms of mechanism it is very similar to the formation of an ester from an alcohol and a carboxylic acid.
Here is the mechanism


text:
What's the monomer in a polycarbonate?
I know what the repeating unit is, but what's the monomer in a polycarbonate?

A polymer name is defined by the repeating unit. Polycarbonates (which is a large group of polymers) further are a nice example where the same repeating unit can be created in a number of ways, from different monomers.  Your question is a bit ambigous with regard to this. Please elaborate. What polycarbonate are we talking about?

Not all polymers are made of single monomers. Polycarbonates usually consist of at least 2 so asking what "the" monomer is isn't quite correct.

Like Dr J said there is no single polycarbonate polymer, the most common one is formed from phosgene (or another carbon dioxide reagent) and bisphenol A. The repeating unit is $\ce{C6H4C(CH3)2C6H4-O-C(=O)-O -}$
In Nagahata, R.; Sugiyama, J.; Goyal, M.; Goto, M.; Asai, M.; Ueda, M.; Takeuchi, K. X-ray analysis of the bisphenol-A-type macrocyclic carbonate trimer. Polym. Adv. Technol. 2000, 11 (6), 294–300. DOI: 10.1002/1099-1581(200006)11:6<294::AID-PAT979>3.0.CO;2-C. (OCAMUA) a tetramer of the repeat unit of this polymer has been reported. This is a small cyclic polymer (oligomer).

There is no one single answer to your question as posed. Polycarbonates (PC) are a group of polymers, the most common of which would probably be the polycarbonate of bisphenol A. The monomer of this PC can be seen here
https://en.wikipedia.org/wiki/Polycarbonate.  There are hundreds of other polycarbonate resins that exist, and they will each have their own unique structure and unique monomer.


text:
Why do two solutes permitted to diffuse simultaneously across a semi-permeable membrane both end up at equal concentrations on both sides?
Let's say we have solute a, which is highly concentrated (let's say 99%?) on side 1 of a semipermeable membrane.
We also have solute b, which is has a much lower concentration (let's say 0.01%) on side 2 of a semipermeable membrane.
If we allow these two solutes to diffuse across a semi-permeable membrane that permits both, my reading seems to suggest that we would eventually expect 50% of solute a to occur on side 1 and 50% on side 2. We would also expect 50% of solute b to occur on side 1 and 50% on side 2.
I'm not sure that I understand this.
I completely agree that 50% of the a+b molecules will end up on side 1 and 50% of a+b will end up on side 2.
But I don't understand why different solutes travel along their own concentration gradients independent of other solutes in solution. Why would they not interact with the other solute?
Perhaps they do, and the diffusion kinetics of solute a would dictate that it reaches equilibrium much faster than solute b?
Would it be correct to predict that the larger the difference in initial global concentration, the longer the lag between when "a" reaches equilibrium vs. when "b" reaches equilibrium?

In a lot of such problems in chemistry the concentration is often used instead of the more appropriate thermodynamic activity. The deviation between activity and concentration tends to become larger with increasing concentration, but there are exceptions. // I think activities are generally ignored because they can't be calculated a priori and using them makes the calculations messy. No problem for a computer doing a numerical analysis, but very difficult to use when doing hand calculations.

This happens because the species do not interact with one another. They just collide with solvent and any other species present (a and or b) and spread out randomly ( 'driven' via entropy of mixing) until equilibrium is established. Then at equilibrium  just as many diffuse left to right as the other way round, so always diffusing. If a precipitate was formed on contact, say, between a and b then one species might all be used up leaving the remainder again free to diffuse until a new equilibrium is established.

If we assume that the solvent is the same on both sides of the membrane (or that the solvent itself can penetrate the membrane) then the final equilibrium will have the same concentration across the membrane.  That only translates to "50% of solute b to occur on side 1 and 50% on side 2" if the volumes are equal.
The rate at which solute A moves from left to right is proportional to the rate at which molecules of solute A on the left side randomly collide with the membrane.  This is proportional to the concentration of A in the left side solution.  The presence of molecules for solute B does not affect this process.
The reverse process is also happening with molecules diffusing from right to left at a rate proportional to their concentration in the right side solution. As the concentration on the right side increases to be equal to the concentration on the left, so the diffusion rates become equal and there is zero nett diffusion and the system approaches equilibrium.
Note that this assumes a "perfect" system where there is no chemical reaction occurring between the solutes or between the solutes and the membrane.  In practice this means that either the interaction between solutes A and B is the same as the interaction between the solutes and the solvent or that the solute molecules are so greatly outnumbered by the solvent molecules that the solute-solute interactions are not significant.
The rate of diffusion of solute A may be different from B (i.e. the proportionality constant between rate and concentration may be different).   This means that before reaching equilibrium the relative concentrations of A and B may change but at equilibrium, the relative concentration will be the same as initially.
If we define "reaching equilibrium" as having some fraction (say 99.99%) of the final concentration then increasing the initial global concentration will increase the lag for both solutes equally and will not change their relative concentrations.


text:
Why potassium chromate can be used as indicator in Mohr method?
Both silver chloride and silver chromate are insoluble, so why is that silver chromate is precipitated before silver chromate? Why wouldn't both red silver chromate and white silver chloride form at the same time?

In the Mohr method for argentometry, the $\ce{[Cl-] > [CrO4^{2-}]}$ so $\ce{AgCl v}$ forms first then as $\ce{Cl-}$ is depleted $\ce{[Cl-] \approx [CrO4^{2-}]}$ and $\ce{Ag2CrO4 v}$ will form as well.


text:
Why is creating water exothermic?
I have to calculate the bond dissociation energy of steam. I'm a total noob, so don't go too hard on me.
(I'm translating the exercise by myself, sorry if there is a mistake.)

Calculate the bond dissociation energy of gas-water/steam ($\ce{H2O}$). Whereas the total enthalpy of the bound is the sum of the "sub-bounds". Use the following table:

Now, we have: $\ce{2H2 + O2 -> 2H2O}$
Imagine we have a tank with one $\ce{O2}$ and two $\ce{H2}$, then, according to the table (if I read it correct), we need $2\cdot436\ \mathrm{kJ} + 497\ \mathrm{kJ}=1396\ \mathrm{kJ}$ to "break" the bonds. We then get two $\ce{O}$ and four $\ce{H}$. They react to two $\ce{H2O}$
Now, according to the table, two $\ce{H2O}$ need $2\cdot 2\cdot 463 \ \mathrm{kJ}=1852\ \mathrm{kJ}$ to bind together.
So, we start with a system of total inner energy $1396\ \rm kJ$ and end up with one of $1852\ \rm kJ$. Clearly, we have to put energy into the first system to get to the second.
Now, that's obviously false - but I don't get why. I'd expect a exothermic reaction because 1. I just know it is and 2. because the $\ce{O2}$ have a double bound which I just expect to be harder to break.
So where is my thought wrong?

Related: Sum of Bond Enthalpies in an Exothermic Reaction

You’re mistakenly conflating the bond dissociation energies between the diatomic components and the product molecule. If the bond dissociation energy of water is larger than the bond dissociation energy of the individual diatomic components, then the system is “lower energy” (preferred) in the $\ce{H2O}$ state and that “extra energy” in the atomic state will be released in the form of exothermic heating. Remember that elemental $\Delta H = 0$, and the water molecule has stronger bonds/lower enthalpy and would require more energy to dissociate than the elemental components (which is what your math shows), meaning water won’t spontaneously dissociate into the diatomic gases since that would be an endothermic process and require some external energy. As an aside, keep in mind that dissociation of bonds in a molecule like water won’t be consistent between atoms (after the first hydrogen bond has broken from an $\ce{H2O}$ molecule, the remaining hydroxyl group requires less energy to dissociate). In sum, if your product has a higher bond dissociation energy than your reactant, it’s at lower enthalpy and the net of the reaction will be a release of energy in the form of exothermic heating.

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

What I do not see in this answer is the calculation of the correct exothermic heating.  You say the HO requires less energy to dissociate into H and O than the H required to separate from the OH, but that would make the total energy to decompose HOH into separate atoms even less than the naïve calculation made by the OP.

Not at all. In these tables, the bond dissociation energy is averaged between atoms of the same element in a molecule like water. For example, the first H atom in water has a bond dissociation energy of approximately 490 kJ/mol, the second H’s BDE is ~425 kJ/mol, and reference tables will give the average of these values so that the calculation of total dissociation into independent elements is still relatively accurate. Here is an example of the exothermic heating calculation.

You say I have to supply $1396 \,\mathrm{kJ}$ to dissociate $2 \ce{H_2}$ and an $\ce{O_2}$.  If I let those separated atoms fall back together as $2 \ce{H_2}$ and an $\ce{O_2}$, do I have to supply another $1396 \,\mathrm{kJ}$ or do I have to transport that much energy away from the system?
It can be helpful to think of a potential energy "landscape".  As we pull the products apart into atoms, we push them up the potential hills (we put energy in).  Then, they fall down the slopes to combine in new molecules (energy comes out).
To form water, we put in $1396 \,\mathrm{kJ}$, then as the atoms fall back together, we extract $1852 \,\mathrm{kJ}$.  So what is the net liberated energy?

EDIT:  You do not "start with a system of total inner energy $1396 \,\mathrm{kJ}$ and end up with one of $1852 \,\mathrm{kJ}$."  The reference point (origin of the coordinate system) of this energy is $0 \,\mathrm{kJ}$, which occurs when all the atoms are separated.  You have to put energy into the $\ce{2H_2}$ and $\ce{O_2}$ to attain this system. so the energy of the initial state is $-1396 \,\mathrm{kJ}$.  That is, you put $1396 \,\mathrm{kJ}$ in just to reach $0\,\mathrm{kJ}$.
The bonded systems don't have "(positive) energy in the bonds" -- they give up energy to be able to form bounds.  Bonds are a form of negative energy.  Exactly the same thing happens with gravity.  Take the Earth-you system.  Right now you are motionless (on an astronomical scale).  We have to put energy into you to reach the zero point of gravitational energy -- infinite separation of you from the Earth.  From any finite separation, if we wait, you will fall back toward the Earth, converting gravitational potential energy to kinetic energy.  When you reach the surface, your gravitational potential energy is negative (since the zero point of that axis corresponds to infinite separation) but your kinetic energy is large.  To recreate the motionlessness that the real you has right now, we have to extract your kinetic energy from the Earth-you system.  That is, starting with zero gravitational potential energy and infinite separation, to recreate what you observe to be the case right now (in real life) we must remove a large amount of energy.
The atoms are in the same condition.  First we have to put energy into the system to pull their bonds apart.  If we were to make them infinitely far apart, we would have to put in (approximately, with details) the amount of energy in your table.  At that point, they have zero chemical potential energy.  Then they fall back together into the molecules you want, converting chemical potential energy into kinetic energy.  Since at the end of the process, we have stationary atoms in our molecules, we have to extract the kinetic energy -- we extract energy from the $0\,\mathrm{kJ}$ state to get the bonded state, so the bonds have negative energy.

"then as the atoms fall back together, we extract 1852 kJ" Thats exactly the point I absolutely don't get. 1. What means "falls back" 2. Why should that "release" energy? HEavier bonds are created, so energy from the system is taken and stored in the bonds. How is that releasing energy?

@xotix : You have a sign flip in your understanding.  Since getting this sign right is very important, I have added some material to my answer.


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Benefits of a Buchner funnel w/ a fritted disc instead of Quantitative Filter Paper Circles
I was wondering if someone could enlighten me on some example us cases where one would be better of using a Buchner funnel with a fritted disc (example), as opposed to a Buchner "style" filter (example) that requires a filter paper (example).
A few times in the past when I was looking to filter a specific substance and asked about fritted discs, most of the replies were "don't use a fritted disc, just use a funnel with the filter papers instead," and this got me wondering when using a fritted disc would be preferable over a filter paper. Originally, in my chemistry naivety, I mentioned that I may end placing a filter paper inside a Buchner funnel with a fritted disc, to "get the best of both worlds" I suppose, but this idea was shot down pretty quickly.
Heres the pros/cons list for both filter types that I have thus far, hoping some can add to it:
Buchner "style" filter with filter papers

Pros

The filter papers are cheaper
Changing the filter paper pore sizes are much easier, as it's just a matter of getting the filter papers with the desired pore size
One can easily cut the filter papers to the proper diameter if needed

Cons

??

Buchner filters with fritted discs

Pros

Filters are reusable if cleaned properly
???

Cons

Fritted discs are much more expensive than filter papers
Fritted discs don't seem to have specific pore sizes
Fritted discs seem like they can be a pain to clean properly

Thanks in advance for any input!

You say "filter papers" but there are Buckner filters made from glass fibers. // A filter is a filter. All sorts of filter materials and filter apparatus types are available. eg Micropore's offerings //  Obviously you don't want filters or filter apparatus for solutions that would be attacked by solution being filtered. For example using a paper filter to filter a conc. sulfuric acid solution would be a bad idea...

@MaxW - I linked to examples of exactly what I was referring to.

Fritted Glass is easy to clean with warm Iso.

Buchner + filter paper cons:

The filtering surface is not uniform. It filters only where the holes are, so most of the surface is not working.
Some solid can slip between the paper and the filter.  Then the filtrate must be filtered again.
You have to cut the paper.

Fritted glass filter pros:

Uniform filtering surface.
Pore sizes from number 1 (coarse) to number 4 (fine). In organic chemistry, number 3 is the standard.
Can be cleaned easily when the solubility of the product is known.
Can be sterilized at high temperature.

I use them almost for everything except for filtering activated carbon and very dirty unknown samples. If the solid does not dissolve in solvents, we try with concentrated caustic, concentrated sulfuric acid or strong oxidants; in that order.
It is better to buy the ones without the vacuum connection.
They are cheaper and have less points where they can break.

easy enough to add a layer of Celite too which will catch very fine particulate. I always use this for catching hydrogenation catalysts on carbon

True.  But if it is not enough or you get a channel it will reach the fritted glass. And then, when you filter a white solid it always get stained.

What if you're filtering a known sample, but its the filtered residue you're after, not the filtrate? And if the residue is in a powdered form? And does it matter if its a base?

Edit: Looks like I could just place a Celite pad on top of the frit, interesting.

Raoul Kessels - If I want to filter out charcoal powder using a fritted disc, as long as I use celite or diatomaceous earth for padding, I should be fine, correct? I know its a bad idea to use charcoal powder with fritted discs (super difficult to clean afterwords), but I think it should be fine if I just create a padded layer. Right?

@Justin Yes, that is how we do it. Make sure you use enough filtering aid and to not disturb the bed when you pour the slurry on it.

@RaoulKessels - Thanks! I just finished a vacuum filtration of a solution containing activated carbon, and the DE definitely helped! What would you say is the best way to clean a frit that was just used with DE for padding? H2SO4/Nochromix?

I have been a synthetic chemist for 10 years and i would strongly recommend to avoid using fritted disks for performing any filtration. In the above answers some points are correct and some others not so much. Therefore i will summarize here:
Filter paper advantages:

Low price.
Variable filtering capacity, up to extra fine by using glass microfiber filter papers (only a couple of times in 10 years i had precipitates that managed to pass a glass microfiber filter paper).
Can be cut to any size and shape.
Can be purchased as circles to sizes that will fit the funnel so cutting is not necessary.
It is disposable, so there is no chance for contamination with material from the previous filtration.
The buchner funnel is easy to clean after use.
Can be easily used with silica or Celite to ensure extremely fine solids dont go through.
Can be sterilized: I have kept normal filter paper in a 180 C oven for several hours. It just turns slightly brown but the filtering capacity doesnt change. Glass micro fiber paper of course can be kept infinitely in the oven without any problems.
Has other uses as well such as folding it for use in funnels for gravity filtration or use in cannula filters.

Filter paper disadvantages:

Can get ripped, and if one scrubs the surface for reclaiming the last trace of solid one can contaminate the product with paper fibers.
As Raoul mentions, some solid can slip between the paper and the filter. However i consider this a mistake form the users side since it wont happen if the filtration is set up properly (had it a few times and it was my mistake in all cases).

Fritted discs advantages:

Less setting up, can be used straight away.
Reusable (?).

Fritted discs disadvantages:

More expensive.
Fragile.
Cant be used for cannula filtrations.
And the biggest issue of all: THEY ARE A NIGHTMARE TO CLEAN. Eventually they will get clogged with all sorts of solids from the various filtrations where one keeps the solution and discards insoluble precipitates. After that point they are practically useless since they could contaminate your filtrate with whatever is stuck in the pores. We had success cleaning blocked frits like that by leaving them in the basebath, but eventually they dissolve making them less effective as filters. The way around this is to use silica or celite but eventually the particles will also get stuck in there (along with whatever else is adsorbed on them). Use of celite also beats the purpose since it is like adding a filter paper to it.

I only use fritted discs (coarse size) in combination with a syringe (had the glass blower fit it in the bottom of the syringe) along with a layer of glass microfiber paper and celite for quick filtrations of small amounts of suspensions. Works like a train toilet.


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What is the softening temperature of styrofoam?
Does anybody know the softening temperature of styrofoam (expanded polystyrene)?
By styrofoam I mean the solid made of expanded polystyrene balls used in packaging.

There are a number of different temperatures one could consider the softening point of atactic (randomly distributed phenyl groups) polystyrene, used in making expanded polystyrene :

Softening point: ~90 °C
Heat Deflection Temperature: 95 °C
Typical Injection Molding Temperature: from 170 °C at back to 230 °C at nozzle

Not to mention softening point is effected on thermal history and residual stress.

And chemical composition (phenyl  distribution etc.)


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Bio dishwashing liquid affecting kitchen pipes?
I have a query regarding bio washing-up liquid.
We've recently experienced slow flow in our kitchen drain pipes, and the plumber whom we asked to have a look at them showed us there was some non-descriptive gunk build-up in the pipes.
We rarely cook with fat (we're vegetarians, and maybe fry a dozen times a year). Plus, we always wipe our pans, pots, and dishes with kitchen paper before doing the dishes in the sink.
We've used bio washing-up liquid for a long time now. The plumber told us that bio washing-up liquids can cause build-up problems in the pipes.
That sounded weird, so we had a quick search on Google, which revealed no information on such issues with bio washing-up liquids.
Would someone be so kind as to let us know whether this is a known issue with bio washing-up liquids?

It might help to add a content list of the washing-up liquid(s)

"Bio" washing-up liquid is a vague marketing term with no specific meaning. Washing-up liquids with such a description in their names could have any of a wide variety of compositions. Without providing more information about the product you are using, it would be impossible to answer. However, it certainly is possible for a cleaning agent to encourage bacterial growth (and potentially formation of a bio-film) if it is /too/ readily biodegradable, acting as a nutrient. This could be made worse by low water temperature, and leaving grey water to stand in the pipes a long time.


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What is the expected product in the reaction of hydrogen chloride with allylbenzene?

Shouldn't it be 4th option since since the carbon attached to the ring is secondary as well as resonance stabilized?

What you say about d is mostly true, but you have to remember to think about the mechanism a little to see which two carbocations you are picking between

I'm confused since I don't know the exact mechanism. Could you please explain me instead of me making absurd guesses?

Ah so it sounds like you should first figure out the mechanism for HCl adding to a double bond, and go from there to solve the problem

I've searched SciFinder for this and related reactions (allyl benzene + H-X). In the results, there are no examples of the halide at the benzylic position in the product, and there are many examples of the halide at the secondary postion arising from the more substituted end of the alkene. One such example is below.

It is certainly warranted to consider both the direct addition and the product from rearrangement (since a resonance stabilized, benzylic carbocation would be formed as an intermediate), and if one were performing the experiment in the lab, it would be a good idea to look for both products. It seems that in practice, the addition of halide to the secondary carbocation is faster than the hydride shift. Predictions about kinetics like this are very difficult, and exam questions that require them could be considered unfair unless there is an overwhelming and predictable rationale for one product over another.

I believe the fourth option would require a removal of a hydrogen.
As well, the double bond is electron donating so it acts as a nucleophile to pick up the proton.
Double bonds like protons/hydrogen more than a Carbon with four single bonds to other carbon and hydrogen. Its happy with a full octet and (nearly) equally electronegative buddies, while double bonds could sure go for a few more protons to fill up.

The fourth would require a hydride shift of the intermediate which is not uncommon.


text:
EQPt pH of a Monoprotic Acid-Base Titration: Product of Molarites Over Their Sum
Context
The Henderson-Hasselbalch equation is as follows. $$pH=-\log\big(K_a\big)+\log\bigg(\frac{[A^-]}{[HA]}\bigg)$$ One may follow its derivation in order to understand how it came to be, yet there are still facets of it that the proof fails to explain. For instance, the equation provides mathematical evidence for why the pH at the midpoint of a monoprotic acid-base titration is equal to the $\ce{pK_a}$. Since $[A^-]=[HA]$ at the midpoint, the fraction reduces to 1. Because $\log(1)=0$, $pH=-\log\big(K_a\big)+0=-\log\big(K_a\big)$, which, of course, gives the pH at the midpoint.
My question is about the same sort of 'insight.'
My Question
Given A mL of B M HA with $\ce{K_a}=C$ titrated by D M xOH at 25oC, where HA is a monoprotic weak acid and xOH is a strong base akin to NaOH, the following expression equates to the pH at the equilibrium point (try it out on Example #1-e here!).
$$pH=14+\log\Bigg(\sqrt{\frac{B*D*10^{-14}}{C(D+B)}}\Bigg)$$
While deriving this, it became necessary for me to determine the $[A^-]$ (it is the major pH determining species at the EQPt). I reasoned as follows.

To find the $[A^-]$, find the mmol of conjugate base and divide it by the total volume.
Since the mmol of conjugate base is equal to the mmol of weak acid used, and the latter quantity is given by the initial volume of acid times its initial molarity ($A*B$), the former quantity is equal to $A*B$.
The total volume, on the other hand, is equal to the initial volume plus what is needed to reach the equivalence point. The initial volume is given by A, so the expression will be $A+V_2$, where $\ce{V2}$ is what’s needed to reach the equivalence point. Take $\ce{M1V1}=\ce{M2V2}$ and plug in the given variables to get $B*A=D*V_2$. Solve for $\ce{V2}$ to get $V_2=\frac{B}{D}*A$. Substitute the value of $\ce{V2}$ into $A+V_2$ to find the total volume, or $A+\frac{B}{D}*A$.
As mentioned before, divide the two expressions to give the final equation, listed below.
$$[A^-]=\frac{A*B}{A+\frac{B}{D}*A}$$

However, the expression simplifies as follows.
\begin{split}
[A^-]&=\frac{A*B}{A+\frac{B}{D}*A}\\
&=\frac{A(B)}{A(1+\frac{B}{D})}\\
&=\frac{B}{1+\frac{B}{D}}\\
&=\frac{B}{\frac{D}{D}+\frac{B}{D}}\\
&=\frac{B}{\frac{D+B}{D}}\\
&=\frac{B*D}{D+B}\\
&=\frac{B*D}{B+D}\\
\end{split}
I find it interesting that such a complicated concept (at least to me) simplifies down to the product of two molarities over their sum. Similar to the way that I justified my initial expression and a facet of the Henderson-Hasselbalch equation, can anyone straight-up explain why this final expression yields the EQPt concentration of conjugate base in the given scenario? If it is possible to keep the answer to a relatively basic, non-organic point of view, that would be much appreciated!
Please note that I have done the best to explain what I am asking -- I think it is a legal but unordinary question for this site. If I can clarify, do not hesitate to comment!

Your insight is a Gordian knot. The Henderson-Hasselbalch equation is thrown around on this site like it is a magic Swiss army knife. Do you understand the implications of using it with a "weak" acid?

@MaxW Probably not, but I’d like to try! Any further clarification?


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Separating specific components of a PUREX raffinate
The PUREX extraction in nuclear fuel reprocessing leaves behind a nitric acid raffinate full of (mainly) 'fission products', which are a diverse set of elements produced by fission in a nuclear reactor. The concentrations of the various components of the raffinate are shown in this image.

My goal is to separate out the group III elements from the group IV+ elements (shown highlighted in red/blue). It does not matter what happens to the rest of the components, they can go with either or neither of these groups in any combination, all that matters is that the group III elements are spatially separated from the group IV+ elements. The process can be as inexact as necessary as long as it mostly accomplishes this goal.
Any thoughts? My nuclear chemistry book says that these group III elements are "have soluble nitrates, chlorides, and sulfates while their fluorides and hydroxides are insoluble. LaF3 is frequently used as a carrier for this group. They form stable complexes with strong chelating agents such as DPTA, EDTA, and the like". Perhaps this makes the question fairly straightforward, I'm coming from a physics, not a chemistry background. I'm just confused as to what properties the group III elements have that the group IV+ elements don't which could be leveraged to achieve some kind of separation.
Sorry for the wall of info, and thanks very much in advance for any help or input!

You have chosen a difficult challenge, a word of advice unless you are working in a nuclear site I would advise you in the strongest terms possible not to attempt a separation of radioactive things in your kitchen, basement, garden shed or some other home lab.
OK warning over, now the idea of adding flouride to create a percipitate of lanthanide (and post plutonium actinides) is a nice idea but it will make any further processing of these elements harder. I also worry that any traces of plutonium or other elements such as zirconium might become incorperated into the lanthanide flouride perceipitate.
I would suggest that you consider solvent extraction as the separation method, I would suggest that you consider the TRPO process chemistry from China. If you were to dilute the PUREX raffinate to reduce the nitric acid content to about 0.5 M, then shake the PUREX raffinate with about 30 % (v/v) of cyanex 923 in an aliphatic kerosene then you will have a organic solution which is a mixture of lanthanides, zirconium, iron and residual plutonium / uranium.
I would suggest stripping (back extracting) the organic phase with strong nitric acid, the plutonium, uranium and zirconium will stay in the organic phase while the lanthanides will go back into the aqueous phase.
In some ways a better alternative would be to use TODGA or a malonamide, a solution of one of these reagents in aliphaltic kerosene will extract the lanthanides from PUREX raffiniate without adjustment of the nitric acid content of the raffinate. You can then strip the lanthanides (plus the americium / curium) from the raffinate. These types of processes have been tested at ITU in their hot cells using PUREX raffinate formed from real fuel.
In some ways I would want to use dilute nitric acid as a stripping agent for the lanthanides from a diamide based process. You can then reextract the lanthanides from the dilute nitric acid with 30 % DEHPA (bis-2-ethylhexyl hydrogen phosphate) in kerosene. In turn back extract this new organic phase with nitric acid, if you are careful it is possible to get some separation of the lanthanides into the heavy and light lanthanides by selective stripping with different concentrations of nitric acid.
Another method which would work would be to use the TRUEX process chemistry from the USA, this is based on CMPO in tributyl phoshate modified kerosene.


text:
Is it environmental to throw ordinary keys in the regular household trash?
I have several keys on my chain that I do not need any more, so I'm thinking about throwing them away, but I'm not sure if it is okay to throw away in the regular household trash. I looked it up to see what are keys made of and I see that it's usually nickel-plated brass, sometimes with silver in the mix (see here and here for example). Is there any environmental harm in throwing away such keys in the trash?
(I am guessing that due to the small size and mass of a key, it would probably not have a significant impact anyway, but I want to know the proper answer nonetheless)

For security reasons I wouldn't throw them in your home trash. Even though the keys may not fit your house, do you want some dumpster diver to try? Slip them quietly into in a public trash can with no ID on the keys. The dump is a witch's stew of nasty stuff. Batteries, electronics with Pb/Sn solder, old cans of paint, and left over bug spray. A metal key is a nothing in comparison.

Environmental is relative. Compared to recycling it is not as environmental but as MaxW said compared to what gets put in the dump everyday its okay.

Depends on where you live. Civilised countries don't just dump their waste in landfills, but sort and treat it, and getting out any solid metal is the first thing to do. It's also profitable.

The question's asking for facts or scientific expertise in helping determine if "there (is) any environmental harm in throwing away such keys in the trash?", which is contrary to the POB close vote.


text:
Diffusion of CO2 and O2 in Water/Atmosphere System
Consider the following experiment:

One reservoir of pure water is interfacing the atmosphere at standard condition.
Both are at rest and at the same temperature.
Both have no internal gradients (ie: constant concentrations, temperature, ...)
The interface between them is a perfect flat surface (for sake of simplicity).
The dissolved $\ce{CO2}$ in the water has a concentration somewhere between 0-50ppm.
The dissolved $\ce{O2}$ in the water has a concentration somewhere between 0-11ppm.

Questions
in order of importance

How to mathematically model the $\ce{CO2}$ and $\ce{O2}$ rate of change per unit area (flux) in the water and in the atmosphere considering only diffusion?
Suppose the atmosphere is at constant velocity relative to the water reservoir. How this would change the mathematical model above?

Notes: Analytical methods would be appreciated instead of 'heavy' computer simulations.

In chemistry the equilibrium conditions are generally what are calculated. The dynamic path to get to equilibrium conditions is much more complex to calculate. Obviously interaction between gas and liquid phases happens at the surface. So there will be some sort of thin layer on gas and liquid side that is not representative of bulk phases which you say are "homogeneously mixed internally." The point is that the "thickness" and profile of the diffusion layers depend on how vigorously the liquid and gas are being mixed. Thus this is a very very complicated problem to handle mathematically.

"In chemistry the equilibrium conditions are generally what are calculated." Ok, I see your point. However, at the moment, I prefer not to focus on the equilibrium.

@MaxW "Thus this is a very very complicated problem to handle mathematically." I imagine someone has already dealt with such problem before (ie: obtain the rate by with gases dissolves into water and vice-versa). Any idea how they might have mathematically model then?

I'll make this last comment.... // Basically you taking about a diffusion problem. But you haven't even supplied details on the physical dimensions of the isolated system. Is this a one inch tube a mile long half filled with water and half filled with gas, or is it a cylinder a yard in diameter but only one inch high? A dynamic model would go on on on in such details.  For an equilibrium calculation they don't matter and you'd just need a few details like temperature, relative volumes, and gas pressure.

@MaxW "But you haven't even supplied details on the physical dimensions of the isolated system." For sake of simplicity, we can consider both to be very large reservoirs (ie: the interface between water and air is an infinite flat surface).

@MaxW "A dynamic model would go on on on in such details." Currently there is no need for a dynamic model. Only the instantaneous rate by which the gases leave the water and go to the atmosphere (and vice-versa).

Such questions are relatively standard in discussing transport phenomena, so I'll suggest Bird, Stewart, and Lightfoot's Transport Phenomena as a good reference.
We assume a flat plane of contact infinite in extent. We assume that the bulk atmosphere has a constant concentration $c_A$ of some species of interest, and that the bulk liquid has a constant concentration $c_B$ of that species. The species is only allowed to mix between the two bulk phases for times $t > 0$. Then the problem reduces to the one-dimensional diffusion equation (Fick's first and second laws) $$\partial_t c(z,t) = D\,\partial_z^2 c(z,t)$$ subject to the initial condition $$c(z,0) = \frac{c_A+c_B}{2} + \frac{c_B-c_A}{2}\text{sgn}(z),$$ and the boundary conditions $$c(-\infty, t) = c_A, \quad c(+\infty, t) = c_B,$$ where $D$ is the species' diffusion constant and $\text{sgn}(z)$ is the signum function.
For ease of solution, make the substitutions $$\bar{c}(z,t) := c(z,t) - \frac{c_A+c_B}{2}, \quad \Delta c := \frac{c_B-c_A}{2}, \quad \phi(z,t) := \frac{\bar{c}(z,t)}{\Delta c},$$ which yields $$\partial_t\phi(z,t) = D\,\partial_z^2 \phi(z,t), \quad \phi(z,0) = \text{sgn}(z), \quad \phi(\pm\infty, t) = \pm1.$$ By definition $\phi$ is a dimensionless quantity, so it must depend on the variables relevant in the problem in a dimensionless form, for otherwise $\phi$ would have units. The relevant variables in our problem are the diffusion constant $D$, the length $z$, and the time $t$.
The only dimensionless variable that arises from this set of quantities is $\eta := z/\sqrt{Dt}$, so we know $\phi(z,t) = \phi(\eta)$. Substituting this result into our diffusion equation and applying the chain rule yields the ordinary differential equation $$-\frac{\eta}{2}\phi'(\eta) = \phi''(\eta).$$ Integrating twice yields the result $$\phi(\eta) = \alpha + \beta\int_0^\eta\text{d}\eta'\,\exp\left(-\frac{\eta'^2}{4}\right),$$ and it remains to determine the constants $\alpha$ and $\beta$ from our boundary conditions. Actually, for standardization, we will first take $\eta' \to 2\eta'$ and $2\eta \to \eta$, so $$\phi(\eta) = \alpha + 2\beta\int_0^\eta\text{d}\eta'\,\exp(-\eta'^2).$$ By symmetry $\alpha = 0$, and by explicit evaluation $\beta = 1/\sqrt{\pi}$, and hence we have $$\phi(\eta) = \text{erf}(\eta), \quad \text{or} \quad \phi(z,t) = \text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ Going back and untangling all our substitutions, we end up with $$c(z,t) = \frac{c_A+c_B}{2}+\frac{c_B-c_A}{2}\text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ The flux then follows as $$J(z,t) = -D\,\partial_zc(z,t) = -D\,\frac{c_B-c_A}{\sqrt{4\pi Dt}}\exp\left(-\frac{z^2}{4Dt}\right),$$ and we are done.

Regarding your second question, I don't believe the movement of one phase relative to the other will have any effect. As we've seen above, the xy-plane essentially drops out of the picture, and you won't be introducing any further gradients in chemical potential by translating one phase relative to the other, since the concentration of each phase is a function of $z$ only. Of course, if you had a finite area of contact, then things would be different. Perhaps see BSL 18.5, in this case.

In addition, keep in mind that there's important chemistry that's being neglected in this simplistic model. For example, the dissolution of carbon dioxide in water follows not just from diffusion but also from reaction with water, forming bicarbonate ion and related species in solution. This model doesn't account for that.

You've correctly understood where I am trying to go, ie, obtain the concentration development with time. However, the focus of my original post was only a small part of your question. Consider the one-dimensional diffusion equation you've written. The diffusion coefficient is constant for z < 0 (inside the water) and constant for z > 0 (inside the atmosphere). But what will be its value for z = 0 (at the interface)? I don't think you can model the interface this way. Fick's equation, in this simplest form, assumes an homogeneous medium, which is not the case at the interface. I guess ...

Math details: your boundary condition c(z,0) is not equal to ca for negatives z values.

"keep in mind that there's important chemistry that's being neglected in this simplistic model." I need to consider this. At least the first order terms. This is another reason why I prefer to focus only on the instantaneous rate of change at t=0 instead of the time development. BTW, this is another reason why I don't feel much confidence on Fick's equations at the interface. There is more stuff going on ...

"not just from diffusion but also from reaction with water" Yeap, I guess you summarize it well. Your one-dimensional diffusion equation is perfectly correct (including at the interface) if we consider only diffusion. However, there is also some chemical reactions going on. I'm not sure whether such chemical reactions would play a major role or not. Any idea how can we estimate the influence of such reactions?

"subject to the initial condition" I guess the correct would be: c(z,0) = ca.(1-H) + cb.H . Agree?

($D$ is not constant over $z$.) Yes, that was quite negligent of me. Luckily the model is still analytically soluble, and details are presented in Sec 3.4 of Crank's The Mathematics of Diffusion. I may transcribe it here if I have time.

(Incorrect initial condition.) Yes, that should be $\text{sgn}(z)$, not $\Theta(z)$. I will edit my answer.

(Instantaneous rate of change at the interface at initial time.) I suspect that flux would not be well-defined because $c(z,t)$ is not differentiable there. I'm also surprised that the initial flux would be sufficient for your modeling purposes.

(Incorporating reactions.) It's possible to incorporate a local equilibrium assumption at the interface and hence to specify the equilibrium constant there, which is also discussed in Sec 3.4 of that book. And yes, I do think it'll play a big role, since I'd expect diffusion to be slower than reaction here. (@MarkMessa)

"I suspect that flux would not be well-defined" Later I've correct myself. What you've done is perfectly fine if we consider only diffusion.

"I'm also surprised that the initial flux would be sufficient for your modeling purposes." If I have the net CO2 and O2 flux across the interface (including diffusion and reactions) as a function of ca and cb, the rest is just a time march procedure that could be easily done in a spreadsheet (although your analytical solution to the differential equations is much better).

Just changed the scope of the OP in order to accept your answer.


text:
% Purity Calculation
An impure sample of calcium oxalate with a mass of 0,803g, is titrated with 15,70cm3 of a 0,101 potassium permanganate
How do I calculate the % purity of the calcium carbonate in the original sample

There seems to be an inconsistency in your question. You presumably want to calculate the % purity of the calcium oxalate, not calcium carbonate.

Find the number of mols that would react with 15.70ml of 0.101 molar potassium permanganate. Then convert it to grams.
Then percentage purity=(x/.803)*100. Where x is the value you calculated above.


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Deciding higher priority order between a triple bond and tertiary carbon
I want to know which of the groups will be given higher priority $\ce{-C#CH}$ or $\ce{-C(CH3)3}$ and why. I tried resolving the triple bond to single bonds such that they are phantom atoms. But that gave me a structure in which there are no atoms after the first point of difference. How is this (see image) wrong and if this is correct then isn't it a contradiction? After first point of difference, there are two difference one is nothing v/s $\ce H$ and other is $\ce C$ vs $\ce H$ and both situation seem to contradict in giving their respective group higher priority order.

I'll point out that there are different schemes for naming compounds. So the answer probably depends on exactly which one that you are using...

@marxw for E and Z or R and S we need to first give them priority  order. And i want to know which priority is correct and why.

At 2nd point of difference, the phantom carbon is attached to the original carbon and hence its a C-C v/s C-H and its obvious that the former will be given higher priority. Its not right to say that the phantom C is not linked to any atom if its really linked to the original C

For a similar question, see this.

From the Wikipedia article on CIP rules.

If an atom A is double-bonded to an atom B, A is treated as being singly bonded to two atoms: B and a "ghost atom" that is a duplicate of B (has the same atomic number) but is not attached to anything except A.

Ghost atoms are attached to resolve the triple/double bond.
So according to CIP rules the two compounds can be resolved as

 The pink atoms are ghost carbons
From here we can see that ethyne has a greater preference over isobutyl as after the first point of difference we move over to the next highest order carbon. In ethyne the second carbon is connected to $\ce{C2H}$ whereas the next highest order carbon in isobutyl is connected to $\ce{H3}$.

What about ethyne vs -C(CH₃)₂-C(CH₃)₂H ?  Does the latter win because they are connected to three hydrogens each, whereas the ghost carbon are not connected to anything?  But then what about... I will draw...


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Can protection group be used in a Wolff–Kishner reaction?
How can I modify the ozonolysis of cyclohexene such that one of the aldehyde groups of the product, adipaldehyde, is efficiently protected as a base stable protecting group? I know that attempting to monofunctionalize adipaldehyde will lead to mixtures.
My aim is to then do a Wolff–Kishner reduction on the monoprotected dialdehyde.

Are you trying to make hexanal from cyclohexene?

@Halake: The protection of adipaldehyde as the monoacetal was solved by Schreiber and Claus some years ago [Also see, Schreiber, S.L.; Claus, R.E.; Reagan, J. Tetrahedron Letters, 1982, 23, 3867.] Ozonolysis of cyclohexene (1) in the presence of methanol leads to the molozonide 2, that opens to the zwitterion 3 which is trapped as the methoxy hydroperoxide 4. Addition of p-toluenesulfonic acid catalyzes the formation of the acetal 5. After neutralization with sodium bicarbonate, reduction of the hydroperoxide with dimethyl sulfide affords the monodimethyl acetal of adipaldehyde 6. Since acetals are base stable, Wolff-Kishner reduction of the aldehyde can be accomplished followed by aqueous acid hydrolysis of the acetal.

I'm assuming you've carried out ozonolysis followed by mild reduction (e.g. using $\ce{Me2S}$) to generate a dialdehyde and now want to convert it to a monoaldehyde. Wolff-Kishner requires basic conditions, so you need a PG insensitive to base. Using ethylene glycol (1 equiv.) in mild acid over a dehydrating agent to generate an acetal would do, but you would need to optimise the conditions to ensure that only one of the aldehyde groups reacts. Then you can carry out Wolff-Kishner as usual and, finally, release the aldehyde in mild acid and excess water.
The question is why would you ever need to do that, if the resulting n-hexanal is readily available from most chemical suppliers?

This seems like an acceptable method, but will definitely yield a mixture of di-, mono-, and unprotected products. Given that the two aldehyde groups are chemically identical, there really isn't a great way to "ensure" only one of them reacts.

Yes, so you’ll probably need to do several rounds of chromatography and recycling of the di and unprotected products to obtain enough of the mono-acetal.

Very much easier to buy hexan-1-ol and do a Swern.

Or buy hexanal already ;)

Under the acetal forming conditions you may also get some hemi-acetal formation just to complicate your chromatography further.


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Where can I get potential energy surface for simple molecules?
I'm looking for potential energy functions for simple molecules, such as water or hydrogen sulfide. I only need the region around the equilibrium geometry, preferably as a Taylor series expanded in terms of normal mode coordinates, with mixed terms given to the fourth or higher order. Just the ground electronic state would do, but ideally I need the first excited state as well.
If these data isn't available, I'd appreciate if you could explain how I could obtain them myself (including software recommendations), given that I'm an undergraduate, who understands the theory but doesn't have much practical experience yet.

psi4 is a very usable, customisable and free/open-source implementation of most mainstream quantum chemistry. You can, with some experience of the syntax in python, have it output what you desire. What level of theory do you want to perform these calculations at?

Extremely accurate state-of-the-art potentials for more than 20 different diatomic molecules (including excited states for many of them!) are given here.
Bob Le Roy has a reputation for building some of the best small molecule potentials in the world, and in recent years he has started to make them available online. But these are not just a Taylor series around equilibrium, they are analytic functions covering the entire landscape of possible geometries, and for diatomics the specific function is given as $V(r)$ here.
You can plug the parameters from the given publications into the $V$ function, and then Taylor expand the global analytic function into a Taylor series if you want, but the original (non-Taylor expanded) function will be more accurate globally.

If the molecule doesn't yet have a potential available, you can build one ab initio (which is just the Latin phrase for "from scratch") using a program such as the free version of MOLPRO which can be found in the row of the linked webpage saying 0.00 EUR.
The input for making potentials in MOLPRO is here.

If you have specific molecules you are interested in, and you can tell us what you wish to do with these potentials, I can try to direct you to the best potentials for the ground and excited states for the particular molecules you ask for. There is no comprehensive database that has all potentials in one place.


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How dangerous is an industrial scale styrene spill?
A few years ago a cisterna truck full of styrene crushed on the road near a small town nearby, spilling 4 tons of the chemical. It apparently has caused a small ecological catastrophe killing fish in the local river with concentrations of 3.7 ml/l in the water, and half of the population - some 2000 people were evacuated in the next 2 days, but it was not mandatory, and what seems more strange to me, I've seen all the firefighters not wearing any gas masks while pouring some white foam on the truck. Also, no one mentioned that styrene is a cancerogen.
Initially, there was no info on the chemical spilled, and I didn't know or care much then, but today I incidentally found out it was styrene, and I got curious. So, were those measures somehow inadequate, or on the contrary - far-fetched, and just how bad this is?

From experience I know firefighters commonly do not take proper precautions and then we are all left to wonder why cancer rates are so high.

Perhaps more to the point, there are any number of chemicals that might generically be called 'styrene' for short. An SDS (Safety Data Sheet) for Styrene (Ethenylbenzene) suggest that it could be pretty bad, particularly through inhalation. It is possibly a carcinogen.

All in all I think that ground contamination would be the main concern now. So if a house near the road where the spill occurred had a water well then that well could become contaminated. Obviously you wouldn't want to have a garden on the spot where the spill occurred either. Driving by the spot wouldn't be a problem.

Yes, it was ethenylbenzene.

Styrene is like mild inconvenience in comparison with lots of really dangerous chemicals.

The question's asking for facts or scientific expertise in helping determine if "there (is) any environmental harm in throwing away such keys in the trash?", which is contrary to the POB close vote

@GaurangTandon Right comment, wrong question.

According to this safety data sheet (SDS), styrene monomer has a Category 3 respiratory classification, where 1 would be most serious. Category 3 is defined "substances produce transient (short duration or temporary) target organ effects."
Add to that these hazard, precautions and carcinogenicity statements:

H335: May cause respiratory irritation.
H372: Causes damage to organs through prolonged or repeated exposure if inhaled.
P260 Do not breathe dust/fume/gas/mist/vapor/spray.
NTP Reasonably anticipated to be a human carcinogen

Read the cited SDS or others for statement on environmental precautions and cleanup. Frankly, they're a bit vague to me, "If the product contaminates rivers and lakes or drains inform respective authorities."

SDS statements seem intentionally vague much of the time. 'Comply with all applicable federal, state, or local laws' seems quite common...


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Why does ethane release more energy than ethyne when burned?
My book explains that since ethane contains a single bond, ethene contains a double bond, and ethyne contains a triple bond, and that when burned ethane releases the most energy because "the fact that the lowest energy is released by ethyne is taken to indicate that this compounds bonds are already in the highest energy configuration. The high energy of  combustion for ethane on the other hand indicates that its bonds have lower energy to start with and can release more energy on combustion." This seems counter intuitive to me. Wouldn't the double and triple bonds release more energy because they're stronger and it takes more energy to break them? Any insight would be appreciated!

Note also that ethane has more hydrogens than ethene and ethyne, which will also contribute to a more negative energy of combustion.

Hi user, welcome to Chem.SE! Please take the tour to familiarize yourself with our site. Thanks!

@a-cyclohexane-molecule has it right. Ethane generates more water. Here is a diagram that helps explain the issue in more detail. The black numbers and arrows are the heats of formation of the three hydrocarbons. Ethane is the only one more stable than the elements from which it is formed. Blue numbers and arrows are the heats of formation of CO2 and H2O from graphite and hydrogen. The red arrows and numbers are the heats of combustion of the three hydrocarbons. Although acetylene and ethylene have a headstart in combustion because the are less stable than ethane, all three compounds produce two CO2's but ethane generates more water than acetylene and ethylene. Ergo, ethane generates the greatest amount of heat. All values are in kcal/mol.

I see, your explanation makes a lot of sense, thank you! I'm still wondering, though, what to make of my book's explanation. How is a lower number of bonds a factor in releasing more energy?

It applies where all other parameters are kept the same. If you were to burn one mole of ethylene plus one mole of H2, the same amount of each species as in one mole of ethane, you'd get more energy than from burning one mole of ethane.


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3,4-Methylenedioxyamphetamine (MDA) basic extraction
At my lab we perform a basic extraction of seized MDA with $\ce{NaOH}$ $\pu{2M}$ prior to GC/MS analysis.  $\ce{MDA + NaOH -> MDA- + Na+ + H2O}$. This reaction should favor the products since $\mathrm{p}K_\mathrm{a}$ of MDA is about 10 and $\mathrm{p}K_\mathrm{a}$ of $\ce{H2O}$ is about 16. All of the MDA should be converted to its ionic form which is soluble in the aqueous layer. Why is there still enough of MDA in the organic layer ($\ce{CHCl3}$) to be detected by GC/MS?

MDA is an alkaloid, i.e., basic. When you talk about its $\mathrm{p}K_\mathrm{a}$ value, you're really talking about the $\mathrm{p}K_\mathrm{a}$ of the protonated ammonium ion.
The ammonium ion is water soluble, allowing us to potentially separate it from the other organic components of the reaction. After making the aqueous solution basic, the ammonium ion is deprotonated to form the free amine, which is much more soluble in the organic phase than the aqueous phase. Then, it is back extracted into clean organic solvent. This isolates (hopefully) just the free amine in the organic layer during the extraction.
So in answer to your question, what you have in the organic layer is the free amine form of MDA, not the ion.


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How are there only 8 structural isomers of C5H11Br?
Everywhere I see mentioned, that there are only 8 structural isomers of $\ce{C5H11Br}$. That is, the simple $\ce{C5}$ straight chain, $\ce{C4}$ with substituted methyl and $\ce{C3}$ with two substituted methyl groups, giving us a total of 8 isomers. But what about substituted ethyl groups? Or substituted bromo ethyl group like $\ce{C2H4Br}$? Am I missing something? Also is there a general formula to know the number of such structural isomers of haloalkanes?

You'll probably need to draw all the possible isomers to see it, but with only 5 carbons to work with the ethyl isomers are redundant with the methyl isomers: i.e. 1-bromo-2-ethyl-propane is the same thing as 1-bromo-2-methyl-butane.
Remember we are only talking about structural isomers, and ignoring stereoisomers.

Okay, I see now. I repeated the same structure in different ways. Is there any short way to determine the number of non-redundant structures?

There's probably some complicated group theory that can, but in general, figuring out the number is too open ended.

Draw n-pentane. 3 places to put Br. Take n-pentane. Clip of a methyl and add it to the remaining n-butane. Only one answer: 2-methyl butane. Add Br to it. 4 monobromo cmpds. Take n-pentane. Clip off 2 methyls and add them to the remaining propane. Only one way, 2,2-dimethylpropane and only one mono bromo cmpd. Does it come out to 8 structural (constitutional isomers? I'm doing this in my head.


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What is the reaction of pentane-2,4-dione with chlorine in presence of NaOH?

We know that in haloform reaction alpha acidic hydrogen are substituted by halogen atom. But in the above case we find two acidic hydrogens in between the two keto groups (due to -R effect) which are more acidic than the hydrogens on the terminal methyl groups.
So what should be the products of this reaction?

That would depend on concentrations - how far reaction will go.

@Mithoron do you have a pdf or any file regarding this. That'll help me. I've googled it but found nothing.

I assume that there is sufficient base and chlorine and that the conditions are vigorous enough to get beyond dichloride 4, which, being a non-enolic β-diketone at C3, is labile toward Haller-Bauer-type cleavage. Acetylacetone 1 is highly enolic and readily deprotonated. The first two chlorinations lead to dichloride 4. Cleavage of dichloride 4 with base gives enolate 6 which can be protonated by water, or to save space, chlorinated to provide the trichloroketone 8, which is an intermediate in the haloform reaction of acetone. Cleavage with base leads to chloroform (10) and a second equivalent of sodium acetate (7).


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Why is sulphuric acid added in the ripper titration?
Ripper titration can be used to determine free $\ce{SO2}$ in wine by titration with iodine.
The method requires sulphuric acid to be added to wine before titration. Why is this the case?

There are 3 forms:

sulfur dioxide (molecular) $$\ce{SO2}$$  ,
Bisulphite $$\ce{HSO3-}$$
Sulphite $$\ce{SO3^2-}$$
(Additionally, bisulphite can become oxidized with water to form sulphuric acid)

These species exist in equilibrium, and shifts in pH with shift the concentration of each species.
The addition of sulphuric acid lowers pH and favors the molecular form. The molecular form can react with iodine and water. When there is no longer any of the molecular form to react with the iodine, the end point its reached and the color changes.
$$\ce{SO2 + I2 + 2H2O → H2SO4 + 2HI}$$


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Difference between reactions of ring strained cycloalkanes and higher cycloalkanes (n>4)
Today I read two publications stating that cyclopropane's ring can be opened by reactions of it with $\ce{Cl2/Br2/HCl/HBr}$. Correct me please if I interpreted something wrong.
Here are these publications:

Lee, C. C.; Hahn, B.; Wan, K.; Woodcock, D. J. Reactions of cyclopropane with hydrochloric acid and of bromocyclopropane with hydrobromic acid. J. Org. Chem. 1969, 34 (10), 3210–3211. DOI: 10.1021/jo01262a097.
Lambert, J. B.; Chelius, E. C.; Schulz, W. J.; Carpenter, N. E. Polar bromination and chlorination of cyclopropane. J. Am. Chem. Soc. 1990, 112 (8), 3156–3162. DOI: 10.1021/ja00164a043.

I read as well that cyclopropane has extra reactivity (due to ring strain obviously) so in the light, it behaves like a linear alkane getting substituted but in the absence of light, ring gets opened (as shown in a ChemGuide article.
Here are my questions about ring-strained cycloalkanes:

can be ring-opening agents for cyclopropane used for cyclobutane?
can we open ring-strained cycloalkanes with hydrogen (and catalyst of course)?
can we substitute cyclobutane like a cyclopropane? (in light)

But, still, I got a few questions about higher cycloalkanes' chemistry

does hydrogen with catalyst open higher cycloalkanes' rings?
in the reaction of $\ce{Cl2/Br2}$ with higher cycloalkanes ($n>4$), ring gets opened or compound is substituted?
do higher cycloalkanes ($n>4$) even react with $\ce{HCl/HBr}$?