How to convert ethanol to benzoic acid? I have tried this method:

Can anyone suggest me a shorter route?

  Criteria: Benzene has to be produced, it is not available.

Edit : I don't understand why the question has been put on hold. They provide the reason that there may be possibly to many good answers or the question is very broad. As far as my knowledge goes, I haven't received any correct answer that addresses what I have asked.

I would request the authority to reconsider on their decision.

Here is how I would do it.

Part I: Blue

First, ethanol 1 is oxidised to acetaldehyde 2, which undergoes nucleophilic addition with $\ce{KCN}$ and $\ce{HCN}$ to form the hydroxynitrile 3, and finally dehydrated and hydrolysed at the same time with sulfuric acid to give acrylic acid 4.

Part II: Red

First, ethanol 1 is reacted to form the Grignard reagent 5, then combines with the aldehyde 2 to form the alcohol 6. It is dehydrated to form butene 7, dihydroxylated to form the diol 8, and dehydrated again to get the diene 9.

Part III: Black

In the final part, 4 and 9 participate in a Diels-Alder reaction with the Lewis acid $\ce{TiCl4}$ to give 10. Selenium dioxide gives the diol 11 and subsequent dehydration gives us benzoic acid, 12.

Although it may be longer, it does not involve things such as high reaction temperatures and therefore may be more feasible in the context of a laboratory.

EDIT: Here is another attempt:

Ethanol 1 is oxidised to acetaldehyde 2, and subsequently with excess formaldehyde and acidic conditions forms pentaerythritol 3, and is oxidised again to form the quadruple aldehyde 4. Strong heating causes 3 to turn into the diol 5. The Grignard reagent 6 is formed and oxidation and reduction of the alkene forms the reactant 7.

Now, both 4 and 7 react to form the ring product 8, and with strong heating thus produces benzoic acid 9.

You Provided a few very great routes I really like the convergent approach of Linus Choy. I'd prefer a shorter route.

In the first step we are preparing the ethylhalogen. There are different procedures. I'd use $\ce{HCl}$ and $\ce{ZnCl2}$. by heating around 60 mins there should be conversion. My Main Point here is so we can go for a Friedel-Crafts reaction.

Second step is the Friedel-Crafts reaction with $\ce{AlCl3}$ or $\ce{FeCl3}$ onto the Benzene Ring. So we end up with ethylbenzene as product.

For the third reaction I also considered $\ce{KMnO4}$ and $\ce{H2SO4}$ for oxidation, there should be procedures leading to the same product.
But I really liked the idea of the oxidation by $\ce{ZnO}$ in DMF here because the reaction only takes a few minutes in the Microwave and has good yields.


What type of acid is found in tea? We were trying the litmus paper test and we found out that tea is acidic. So the question came in my mind and my science teacher could not give me any satisfactory answer: what acids are found in tea?

Some of the major components of tea are polyphenols. They are fairly acidic and may dominate the pH of tea.

For some examples of the types of polyphenol that are common see this site on tea chemistry:

  The major flavanols [a specific type of polyphenolic compounds] in tea are: catechin (C), epicatechin (EC), epicatechin gallate (ECG), gallocatechin (GC), epigallocatechin (EGC), and epigallocatechin gallate (EGCG).

EGCC has this structure:

from which some degree of acidity is likely to result from all those phenolic OH groups.

Tea has Tartaric acid as tea is a little bit bitter in taste. Also many other foods have tartaric acid such as grapes and lemon.


Turn Alpha Helix into Beta Pleat Beta sheets have the peptide backbone parallel whilst the alpha helix has a curled backbone.
But the thing I can't get around my head is why would the same protein always fold into one of these shapes?
It looks to me like I could grab an alpha helix and just straighten it out to form a beta pleat. I can't see anything that wouldn't work - so if I could turn alpha into beta, then what ensures that these proteins will fold the same way each time they're made? I think there must be something that prevents me from turning alpha into beta but I'm not sure what.

In the alpha helix structure as well as the beta pleated structure, the atoms are arranged in such a way that the number of hydrogen bonds is maximum. H bonding is a stabilizing factor, and consequently, any molecule (or polymer) would like to maximize their number. This answers your question as to why the protein would fold into one of these shapes.
Coming to why alpha and beta structures are not easily interconvertible, it is once again because of hydrogen bonding. The H bonds between H and electronegative atoms like oxygen and nitrogen are difficult to break, and hence we cannot simply smash an alpha helix into a beta pleated structure, since in doing so, we would inevitably have to break some hydrogen bonds. Thus, whether a protein exists in alpha or beta form depends on how it was formed.


Conversion of a value between g/gmol and lb/lbmol I am doing some air emissions calculations and I really just have a bad feeling about my results.  By definition, molecular weights are given in $\pu{g/gmol}$.  For the substance I have, I was told by a technical contact at Citgo that this material has a molecular weight of $\pu{140 g/mol}$ (which should be $\pu{g/gmol}$).  Whenever I convert this $\pu{lb/lbmol}$ (which is required for this particular equation), I'm getting a value of $\pu{0.3086 lb/lbmol}$.  There are 453 grams in a pound.  Is it not correct to divide 140 grams by 453 grams to get the total number of pounds?

My final result for annual emissions is pretty low.

A gram-mole ($\pu{gmol}$),often just called a mole ($\pu{mol}$), is the amount of a material whose mass in grams equal to the numerical value of the atomic or formula mass. For example, a gram-mole of water has a mass of 18.017 grams, this number being the formula mass for water.

A pound-mole ($\pu{lbmol}$) is the amount of a substance whose mass in pounds-mass equals the atomic or formula mass. For water, one pound mole has a mass of $18.017$ lb-mass.

The pound-mole is bigger than the gram-mole by the same factor as a pound is bigger than a gram - so pounds per pound-mole equals grams per gram-mole.  You do not divide by anything.

https://www.calculand.com/unit-converter/?gruppe=Molar+Mass&einheit=Pound+per+pound-mole+%5Blb%2Flbmol%5D

1 lb / lb-mol is exactly 1 g / mol, because 1 lb-mol is 453,59 mol, which is the same ratio that there is between lb and g. The link above is a very useful converter.


Interpretation of the following mass spectrum I'm given the above spectrum. $M^+ = 102$, probably $\ce{C_6H_14O}$ or $\ce{C_5H_10O_2}$.
I'm thinking of ethyl propionate. (that picture also shows the fragments I expect). I can account for the following peaks:
m/z = 73: Loss of ethyl from acyl side (expected lowish intensity)
m/z = 57: Loss of alkoxy to form $\ce{CH3CH2CO+}$ (moderate intensity)
m/z = 45: Alkoxy ion (weak)
m/z = 29: ethyl carbocation, moderate intensity.
I can't account for 74, since according to my book the alkoxy side would not proceed through normal Mclafferty, but the Mclafferty + 1 rearrangement... That should yield m/z=75.
I proposed methyl butyrate, and of course that's not right. My instructor wrote to pay attention to the m/z = 57 (M-45) and m/z = 85 (M-17). This fits well with carboxylic acids, but again cannot account for the m/z = 74. McLafferty of carboxylic acids would result in a fragment of m/z=60...
Can someone help me out, I've come to a dead end for hours now.

I was on to the wrong structure. My main issue with the carboxylic acid was the fragment m/z=60 that it would produce by McLafferty. I didn't have that, so I thought that left me with t-Bu-COOH, which was the only structure with no $\gamma$-hydrogens. Besides I couldn't explain the m/z=74 peak with an acid. On the other hand I couldn't explain the M-17 peak by ethylpropanoate.

But of course, if the carboxylic acid has a methyl substituent at the $\alpha$-carbon, then that methyl group would also be part of the McLafferty fragment, therefore resulting in m/z=74!

The spectre is of: 2-methyl-butanoic acid

M-17 and M-45:

Alpha cleavage of carbonyl => loss of hydroxyl radical or carboxyl radical

m/z = 74:

Loss of ethene by McLafferty rearrangement

m/z = 45:

$\ce{COOH^+}$ fragment

M-15 and M-29:

Loss of methyl radical or ethyl radical

m/z = 29:

Ethyl carbocation

For m/z 74 it could be methyl ester:  in the loss of a $\ce{CH2=CH2}$ by McLafferty rearrangement.
$\ce{CH2=C(OH)OCH3+}$


Identity a metal based on its properties A metal X can be cut with a knife. When placed in water, metal X floats and melts. What is X? The options are calcium, copper, magnesium and potassium.

I think it can't be potassium because it will react vigorously with water, So I believe it's calcium. But the only problem is that calcium is denser than water so it will sink.

Copper is much denser than water and it does not react with it. Magnesium is also denser than water and reacts only very slowly. Calcium will also sink in water, although it will react with it producing hydrogen, so small pieces of the element may float on the surface. However, it would definitely not melt. The correct answer is potassium - it is a very soft metal, less dense than water and reacts with it producing enough heat to melt it (potassium melts at only $\pu{63.5^\circ C}$).

Calcium has the lightest density of all the earth's alkali metals. It is so soft that calcium can be cut with a knife, yet it can be harder than lead. Unlike magnesium, calcium is not easy to ignite.  so your answer is correct


Why does K have a greater electron affinity than Ca? Why will $\ce{K}$ have a greater electron affinity than $\ce{Ca}$? I know that $\ce{Ca}$ has a greater first ionization energy, and I also know that $\ce{K}$ has a greater second ionization energy, but I am confused on how to explain this.

Neither K nor Ca "wants" to ionize.  Both ionization energies are endothermic.  This question refers to eelctron affinity, which is gaining an electron.

K has a more exothermic electron affinity because the electron gained fills the 4s orbital.  In Ca the electron goes to 3d.  The energy gap between 4s and 3d is enough to make the process barely exothermic.

Note that the numbers given in tables are positive but that the process is exothermic.

the octet rule. All elements want 8 electrons in their outer shell. K normally has only one electron in its outer shell and removing it exposes the full octet shell underneath which is favorable and spontaneously occurs, which is why K exists as K+ in solution. So K actually wants to ionize to +1 but no further. However Ca has two electrons in its outer electron shell. Therefore it wants to loose two and expose the next full shell of eight electrons, which is comparable to the electron configuration of Argon (Ar).


Assigning locants in halo alkanes/alkenes In the first example, if you add up the number of carbon which has a substituent, you will get 2+4+4 = 10. However, if you start from left to right, then you'd get 2+2+4=8. Why don't we follow the latter one when it gives us a lower number? (Lowest set of locant rule.)

The same applies for the second problem: we could've gotten 2+3+4, but we chose 2+4+5, why so?

Indeed, the first compound in your sketch is numbered incorrectly according to IUPAC. IUPAC rules for branched hydrocarbons say: "When series of locants containing the same number of terms are compared term by term, that series is "lowest" which contains the lowest number on the occasion of the first difference. This principle is applied irrespective of the nature of the substituents. " (emphasis mine). In your top case, this means that you have two numbering choices: either 2,2,4 or 2,4,4. The 2,2,4 choice has a lower locant at the first point of difference, so it is preferred. When you construct the name, you indicate the substituents in alphabetical order, so the full name of the top compound is 4-chloro-2,2-dimethylpentane.

If unsaturations are present, then these have a higher priority than alkyl and halogen substituents. From the rules for unsaturated hydrocarbons: "The chain is so numbered as to give the lowest possible numbers to the double bonds." This rule is sufficient to impose the numbering you report.

As Loong correctly pointed out, the links provided above refer to the text of the now obsolete 1979 recommendations. Both the 1979 and 1993 recommendations are available on the same site, maintained by ACD Labs with IUPAC's permission, and are in accordance with what shown above.

Unfortunately, the text of the most recent (2013) recommendations is not freely accessible (that I know of). This is a link to the electronic version of the book, published by The Royal Society of Chemistry. The relevant section is P-14.3.5 in Chapter P-1 (General Principles, Rules, and Conventions): "The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set '2,3,5,8' is lower than '3,4,6,8' and '2,4,5,7'."

First point of difference is applied here. Chloro comes first in alphabetical order and lowest sum rule is applied when first point of difference is not working.
In 2nd case double bond has higher priority over substituent.


Why is condensation in dyed beverages colorless? Does food coloring somehow evaporate and turn into nothing?   Or does the beverage itself being much more color intensive?

I had a bottle of Gatorade, colored Lemon-Lime (and flavored as such). I noticed that after leaving the bottle in the car during a hot day and returned to the car, that condensation had formed on the top parts of the bottle.

But the condensation is colorless; does this mean it is solely water that is evaporating from the drink itself as condensation occurs, leaving food coloring or the color of the beverage behind?

Given that the majority of beverages such as Gatorade is water, as stated in the comments, the condensate is predominantly water, as can be seen in the Patent Electrolyte blends providing reduced salty taste (which includes Gatorade), that water amounts to well over 99% by composition, of course this will vary. The dyes themselves don't make up much of the volume.

A common food dye used, according to the Indiana University online experiment webpage How Much Red Dye is in My Gatorade®? , is the red dye, also known as 'Allura Red'

its properties according to the Wikipedia page are stated as:

  It usually comes as a sodium salt, but can also be used as both calcium and potassium salts. It is soluble in water; Its melting point is above 300 °C (572 °F).

It is likely that the majority of the food dye based salts would remain in solution.

Any salts (such as the example above), transported with the evaporating/condensing water would be too small of a concentration to cause any visible effects.

A similar question and answer is Does salt vaporise?

I thought the answer was that condensation is on the outside. As such, it's the colorless water from the air, not the Gatorade, that lacks color.


Predict which hydrogen is more acidic in the following organic compounds? Between 2-chloro-2-phenylpropionic acid and hydrogen phthalate, which is the more acidic compound?

I'd say the $\ce{H}$ on the second one is more acidic as the $\ce{-COOH}$ group is directly attached to the aromatic group. Is that correct?

In the second molecule, the charge on the carboxylate ion is not stabilised through resonance, so we can rule out any resonance stabilisation. Moreover, the fact that the benzene ring is in fact electron donating within the $\pi$ system doesn't help either.

More importantly, however, (as it was pointed out to me by @orthocresol and @hippalelectryon in chat) one can expect intramolecular hydrogen bonding.

Moreover, after the second deprotonation, the second molecule would have two negative charges in close proximity to each other, and this, again, is obviously destabilising.

A combination of these factors would make the second molecule a weaker acid.

In the first case, however, the inductive electron withdrawing effect would increase the acidity, (and the benzene ring is far enough to not have an effect).

Thus, I would say it is the first compound that is more acidic.

@Hippalectryon used a software to compute a pKa of 3.87 for the first one, and Wikipedia records 5.51 for the second one, which confirms my claim.

@Lydia Hernandez. 2-chloro-2-phenylpropionic acid is certainly more acidic than hydrogen phthalate. The reason is very simple, A -COOH group has a well defined conjugated system of its own. It is because of this conjugated system the acidic behavior of this group is defined. On the other hand whenever a -COOH group is attached to the benzene ring directly it (benzene ring) disturbs the conjugation of -COOH group, thus the leaving of H becomes difficult. This disturbance is known as "cross Conjugation". You can observe the same with the help of following figures -

You can verify the answer via pH or pKa values.


Can flames be colored black? Most of us here would know that flames can be colored by adding an appropriate salt to the 'fuel' or by simply introducing it into a flame. Boron and Barium salts give the flame a green tinge, Strontium colors it red, Sodium does gold, copper a bluish-green, etc, etc.
Now from my (not-so-in-depth) high-schooler level of understanding, the energy released by whatever is being burnt to produce the flame in the first place (as is usually the case), excites electrons in metal ions in the salt to higher energy levels. And when these electrons 'de-excite' and drop back to the original energy levels, they emit that extra energy as photons (whose frequencies are determined by the Planck equation $E = h\nu$ ) and these photons are what we perceive as the colored flame.
If I'm not mistaken the book Harry Potter and the Philosopher's stone mentions a 'black flame' of sorts, being one of the obstacles Harry has to circumvent in order to reach the Stone. (Artist's depiction)

Now, as I understand it, for something to be 'black', it must absorb all ( or at least a fairly significant amount of) visible light, encompassing a wide range of frequencies.
Going back to colored flames, I mentioned that the ions must emit light energy, not absorb it, in order to color the flame.
Armed with all this information, would anyone know if it really is possible to create 'black colored flames'? Has it ever been achieved?
[I tried thinking it out myself.....just gave me a headache]
Edit 1: And yes, I've made note of what @Mithoron mentioned, but what I mentioned was just conventional logic, it isn't universal though.
...Newtonian mechanics was 'conventional logic', but we all saw how that failed in relativistic scenarios. So I didn't want to take the risk, hence I've asked If it actually is possible...
Edit 2: Just because someone mentioned otherwise in an answer... I emphasise that by black flames I don't mean invisible flames.

This video shows how a black flame is achieved. If you illuminate the fire with a monochromatic light source (sodium vapor lamp) and introduce a species in the fire that absorbs that wavelength (sodium ions) then the fire will in fact appear black under the illumination. A screen shot is shown below:

Click image for video.

Using an (I think) hydrogen vapor lamp and salt water it is possible to make the fire absorb all of the light at the particular wavelength that the vapor lamp emits.


Why does bromine add to a double bond instead of a triple bond if both are present? According to my textbook, the addition of bromine to pent-1-en-4-yne at low temperature yields 4,5-dibromopent-1-yne:

Why does bromination of the double bond occur selectively?

I'm going to take "more reactive" as in the alkene undergoes bromination at a faster rate than the alkyne, which seems to imply that the activation energy of the transition state is greater.

A Kinetic Perspective

The formation of the bromonium ion involves a cyclic transition state where the the Br-Br bond is attacked by the alkene/alkyne pi-bond, and the pi-bond donates to the sigma* antibonding orbital to form a sigma C-Br bond. In doing so, both carbons in the alkene/alkyne will have their p-orbitals hybridized with the sp/sp2 orbital to form the sp2/sp3 orbital respectively. However, the sp2 bonding orbital will have more ring strain, as it naturally exists with bond angles of 120 degrees away from other sp2 orbitals, while the sp3 orbital is somewhat less strained with bond angles ~109.5 degrees apart. Thus the higher energy transition state will make the alkyne less reactive, and bromination occurs on the alkene first.

In addition, (as was pointed out here : https://www.masterorganicchemistry.com/2013/05/29/alkyne-addition-pathways-the-3-membered-ring-pathway/) the p orbitals on Cl conjugate to the pi-bond on the alkene transition state, resulting in a 4-electron anti-aromatic system! This further increases the instability of the transition state and results in a slower reaction.

TL;DR: There's more ring strain on the alkyne's transition state because the carbons are sp2 hybridized, and its transition state involves the formation of an anti-aromatic ring, so the reaction occurs slower than bromination on the alkene.

In general, halo addition is an 'Electrophillic' addition, not a 'Nucleophillic' addition and hence it is $\ce{Br+}$ which attacks first on the saturated hydrocarbon.

Now coming to your question,
According to Jerry March (4th edition, Pg748) Advanced Organic Chemistry,

  There is a higher concentration of electrons between the
  carbons of the triple bond than in a double bond, and yet triple
  bonds are less subject to electrophilic attack and more subject
  to nucleophilic attack than double bond.

The probable cause, as suggested by Sujith seems a pretty reasonable explanation backed up by the fact that while additions with 'bridged' intermediates react faster with double bonds, additions without such intermediates react almost at the same rate with both entities.


Why does mercury not form amalgam with iron? I am learning about solutions and while giving an example of a liquid (solute) and a solid (solvent), my teacher told us that mercury forms an amalgam with all the metals except iron.

I want to know why this is so?

Iron does not form an amalgam readily, hence the use of iron electrodes in mercury switches.  Apparently, iron does form alloys with mercury under specific conditions; see http://www.madsci.org/posts/archives/2011-04/1304143502.Ch.r.html.

By contrast, mercury spilled on gold (e.g. gold-alloy jewelry) soaks in immediately, discoloring the gold, and migrates further into the bulk metal in a few hours, thus restoring the color.

In Fe,Co,Ni The Metallic Bonds Are Very Strong Hence Don't Allow Mercury To Diffuse Into It Hence No Amalgam With Them While In Other Metals Like Ag,Au,Na etc(Soft Metals) Mercury Can Easily Diffuse And Form Metallic Bonds With Them(Ag,Au,Na,etc) Hence Amalgam Formation In Them.
Note:-All Metal Mentioned Above Are Of Comparable Size Of Hg i.e. Fulfilling Amalgam formation Condition.


Help balancing reaction between Iron (II) chloride and potassium phosphate Very rusty with chemistry and looking to review some basics. Came across the following:

  Iron(II) chloride and potassium phosphate react. Write and balance the molecular, ionic, and net ionic equations.

What I have so far (guessing, basically):

$$\ce{FeCl2 + K(H2PO4) -> KCl + Fe(H2PO4)2}$$   [still need to balance but correct right hand side?]

I can't remember the exact rules for forming the right hand side. I just put this in place to match ion charges. What prevents me from writing the right hand side as something else entirely (not sure what but I'm asking, "well, couldn't I figure something out if I break everything apart and re-arrange into another option?").

EDIT: Additional Question Below

Ok, so assume the above reaction was balanced (it's not) and valid. Then, my question is does the reaction go both ways ($\ce{<-->}$)? If so, when/why/how? I'm trying to think back many years ago and it has something to do with "reactivity" and the row of the elements on the periodic table, correct? I'm going to look into it more myself but if someone has a quick "30 second" answer I'm sure it'd be of help to me!

[Assuming this is OK and valid in the forward direction. Again, needs to be balanced.]

$$\ce{FeCl2 + K(H2PO4) -> KCl + Fe(H2PO4)2}$$

[Is this ever ok/possible?]

$$\ce{KCl + Fe(H2PO4)2 -> FeCl2 + K(H2PO4)}$$

This question was the continuation of your question and since both question are related,  I am answering the question by emphasizing the mistake you are doing and adding a relevant portion to explain it.

Mistake:

Iron(II) chloride and potassium phosphate react. Write and balance the molecular, ionic, and net ionic equations.

You need to consider potassium phosphate to be $\ce{K3PO4}$ not $\ce{KH2PO4}$ according to IUPAC's Red book (see sections IR-8.2 and IR-8.4 and the anions in question are explicitly named in table IR-8.1 therein.)

If $\ce{KH2PO4}$ is meant to be added, the correct term should be potassium dihydrogenphosphate.

Extra info.:

Iron hydrogen phosphate are new in this world and has been recently synthesised and as not as stable as iron phosphates. So, only iron (II and  III) phosphate and potassium chloride forms from the reaction.

$$\ce{3FeCl2 + 2K3PO4 ⟶ Fe3(PO4)2 + 6KCl}$$

$$\ce{Fe3(PO4)2 ->[air oxidn⋅,400C] FePO4}$$

or actually

$$\ce{4Fe3(PO4)2 + 3O2 ->[400 C] 8FePO4 + 2Fe2O3}$$

Iron hydrogen phosphate can be made by reacting iron(II) phosphate with dilute hydrochloric acid.

$$\ce{Fe3(PO4)2 + 2HCl → 2FeHPO4 + FeCl2}$$

The molecular formula: $$\ce{ 2 K3PO4 (aq) + 3Ni(NO3)2 (aq) -> Ni3(PO4)2 (s) + 6KNO3 (aq) }$$
The complete ionic formula is: $$ \ce{6K+ (aq) + 2PO4^3- (aq) + 3Ni^2+ (aq) +2NO3-  -> Ni3(PO4)2 (s) + 6K^+(aq) + 2NO3^-(aq) }$$
The net ionic formula is: $$ \ce{2PO4^3- (aq) + 3Ni^2+ (aq) -> Ni3(PO4)2 (s) }$$


How does the presence of chlorine atoms in PVC make it flame resistant? In a paper that I'm reading, it states that:

  The presence of chlorine atoms in polyvinyl chloride make it flame resistant

and I'm just wondering why the presence of chlorine atoms will make it flame resistant. To the best of my knowledge, burning something is just providing a material with enough activation energy such that it can react with oxygen to produce products such as carbon dioxide and water. If so why does the chlorine prevent this process in any way?

PVC is not truly flame resistant, it does decompose on relatively gentle heating (around 150 °C) However, the process produces significant volumes of hydrogen chloride, that is not flammable and isolates the plastic from the flame. This makes makes PVC self-extinguishing or fire-retardant.

My intuitive answer is that Cl has a relatively high electronegativity, so it pulls electrons towards itself, making it harder for the oxygen to react with the molecule. Reactions generally have an oxidizer which attracts electrons, and a reducer that supplies electrons. Oxidation of an organic compound usually means the carbons supply electrons to incoming oxygen atoms, making polarized and thus stable bonds. The bonds with the Chlorine atoms are themselves polarized, making the chlorine somewhat negative and the rest of the molecule slightly positive, making it less eager to give up electrons to an oxygen atom, meaning a resulting bond with oxygen would be less polarized and less stable....


How to find pH by mixing two solutions of different concentrations? A mixture is made by combining 110 mL of 0.15 M $\ce{HCl}$ and 215 mL of 0.055 M $\ce{HI}$.  What is the pH of the solution?

$$\ce{HCl -> H+ + Cl-}$$
$$\ce{HI -> H+ + I-}$$
According to what I thought, the molarity of $\ce{H+}$ is the same as $\ce{HCl}$, because it is a strong acid and the mole ratio.  So, I added the molarity of both acids: $\pu{0.15M} + \pu{0.055M} = \pu{0.205M}$ for $\ce{H+}$ in solution.

$$\mathrm{pH} = -\log[\ce{H+}]$$
$$\mathrm{pH} = -\log[0.205]$$
$$\mathrm{pH} = 0.689$$

You are right in assuming the two acids are strong acids and therefore completely dissociated, but you can easily look that also up, e.g. here.
Therefore you can assume, that the concentration of protons is equal to the concentration of the acid.
\begin{align}
c_{\ce{HI}}(\ce{H+}) &=  c(\ce{HI}) &
c_{\ce{HCl}}(\ce{H+}) &=  c(\ce{HCl}) &
\end{align}

Now you need to determine the amount of the protons in each solution with $$n=c\cdot V.$$

You have to add the amount of protons in each solution and determine the combined volume.
\begin{align}
n_\mathrm{tot}(\ce{H+}) &= n_{\ce{HI}}(\ce{H+}) + n_{\ce{HCl}}(\ce{H+})&
V_\mathrm{tot} &= V(\ce{HI}) + V(\ce{HCl})
\end{align}

To calculate the pH of the resulting concentration, just take the negative logarithm.
$$\mathrm{pH} = -\log c_\mathrm{tot}(\ce{H+}) = -\log\left[\frac{n_\mathrm{tot}(\ce{H+})}{V_\mathrm{tot}}\right]$$

   Your solution should be $\mathrm{pH}=1.06$, rounded to three significant figures.

The approach you chose can only work, if you have equal volumes of the two solutions.


Will mixing hydrogen and oxygen without a spark result in combustion? I plan to do the electrolysis of dilute sulphuric acid. This will produce hydrogen gas and oxygen gas at the anodes. I do not plan to collect these gases, but to simply let it form and release into the air without separating them. Would this carry any risk whatsoever, considering that an electric setup is nearby?

For hydrogen to burn or explode in air, its concentration needs to be at least 4%.

Unless you're using very high power levels, you're unlikely to reach that concentration in the room's air. You could ignite bubbles of hydrogen as soon as they pop at the surface, but if you don't, the hydrogen (with about 1/8 the density of air) will rise quickly, diffusing into the air as it goes.

If you're doing this using a battery, or a small line-operated power supply, there's very little danger of explosion.

I asked a somewhat similar question a few weeks ago and Poutnik made a smart suggestion that I simply mix the H2 with nitrogen to keep it safely below explosive concentrations, which I set out to do prior to extracting it outside with a fan. So the nitrogen ensure that concentrations are low enough for the fan engine's sparks not to light it on, and the fan extraction avoids both hydrogen and nitrogen accumulation in the building.

You can buy the fans and tubing in stores made for hydroponic gardening. The tubing looks like this:

Also keep in mind that every time you turn an electric switch on or off, a voltage peak occurs which ionises a small amount of air between the contacts and creates a tiny spark, so an electric setup nearby that is turned on / off may be hard to separate from having a "spark", which may be as dangerous as any other sparks if the gaseous mix penetrates the switch area.

Sources

How can I neutralise H2 gas without burning it?
https://www.quora.com/What-causes-sparking-in-the-switches-when-light-is-put-off


Why are magnesium chloride and calcium chloride more soluble than sodium chloride? I read that $\ce{MgCl2}$ and $\ce{CaCl2}$ are more soluble than $\ce{NaCl}$ in water. Solubility of $\ce{MgCl2}$ is $\pu{543 g/L}$ and that of $\ce{NaCl}$ is $\pu{360 g/L}$ (both at $20^{\circ} \pu{C}$).

I think that $\ce{NaCl}$ should be more soluble due to its higher ionic nature. $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ are more polarizing and have more covalent nature and thus should be less soluble.

I want to know the reason for why the opposite is happening.

I don't think there is any simple answer.There are some rules of thumb, all of which have exceptions. Most chlorides are soluble. That includes both MgCl2 and CaCl2.

Solubility is the result of the balance between competing interactions: between the ions in the salt and between those ions and the water.

Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.

The longer answer is to do with something called the "enthalpy of solvation".'

you can think it this way that the lattice enthalpy in MgCl2 and CaCl2 will be less and can be easily be overpowered by hydration enthalpy

$\ce{MgCl2}$ and $\ce{CaCl2}$ are more soluble than $\ce{NaCl}$ because In $\ce{MgCl2}$ and $\ce{CaCl2}$ the ionic charge is +2 hence it can interact with more number of $\ce{H2O}$ molecules, as the number of interactions increases more amount of energy is liberated which is a hydration energy which is sufficient enough to overcome the lattice energy of the compound.
Even solubility increases with increase in charge on ion.


Mechanism of Diels-Alder reaction I have a question about the mechanism of the Diels-Alder reaction. Is this a two-electron or one-electron process as described by Fig (a) or (b) respectively or does not matter? How do we know that?

First of all, full arrow represents a movement of 2 electrons. Half arrows represent movement of 1 electron.

Diels-Alder is a 6π electron pericyclic reaction. The diene provides 4π electrons and the dienophile 2π electrons to the formation of the bonds. The Diels-Alder reaction belongs to [4+2] cycloadditions according to the Woodward-Hoffmann nomenclature.

A purely formal way of determining the participating electrons consists of simply counting the electron flow arrows that are used to illustrate the mechanism and multiplying by two. Each arrow moves two electrons.
You have 3 arrows representing a reaction, so 3*2=6 electrons

The number of electrons participating in the cycloaddition determines wether the reaction follows either a supra-supra (s+s) or a supra-antara (s+a) mechanism.

This is a nice source for Diels-Alder mechanisms


Are all complexes with a polydentate ligand examples of chelation? I'm getting a little confused about the definition of chelation and its implications (I should probably point out I'm only an A2 Chemistry student). The IUPAC defines chelation as the following:

  The formation or presence of two or more separate coordinate bonds between a polydentate (multiple bonded) ligand and a single central atom

Does this mean that any ligand that is not monodentate (e.g. even ethanedioate) is an example of chelation? I thought chelation was only for complexes with a large coordination number (e.g. $\ce{EDTA^4-}$), it would seem odd to call a single metal ion with 1 ethanedioate ligand chelated?

I think you should not focus on the ligand to define chelation. You should instead focus on the complex that is formed. In fact, the very same multidentate ligand may participate in binding in more than one way:

by forming more than one bond with the same central atom, therefore forming a chelated complex;
by forming only one bond with that central ion, even though it could form two; this may be the case when a vast excess of this ligand is present with respect to the metal ion;
by bridging between two adjacent central ions.

Only the first of these possibility is referred to as chelation. In other words, forming a chelated complex doesn't depend solely on the fact that the ligand is multidentate. Being multidentate is a necessary, but not sufficient condition for a ligand to form chelated complexes.

Chelation is simply foratiom of a complex by a ligand which has two or more then two sites within a molecule,the word is derived from crabs claw as it grips the central metal atom and all polydentate ligands can act as chelating agents...


What's the actual or official definition of activation energy? In some sources I have seen that activation energy is the minimum energy required to cause a reaction. This implies that activation energy doesn't change with temperature (not considering very high changes).
In other sources it is the difference between the energy required to cause the reaction & the energy of the reactants. This implies that activation energy is temperature dependent since the energy of the reactants change with temperature. So which is correct?
Also another source (khan academy's guardian chemist) says that "The official definition of activation energy is a bit complicated and involves some calculus."

Activation Energy is the minimum amount of energy needed extra than the average potential energy of the reactants to cause a reaction.And it does not change with the change of temperature, and for clarification the energy of the reactant molecules don't also change with the reaction. It is just the rate which changes with the reaction according to the arrhenius equation k=Ae^{-\frac{Ea}{RT}}So,the fraction of molecules having energy higher than activation energy changes with change in temperature. I think the easier definition can be stated like this


How to replicate compositions of discontinued paints? I am a house painter, over the last 20 years I've noticed that discontinued lines of paint are much better. These paints were vinyl acrylic. Now the vinyl has been removed. I'm wondering if adding 1 ounce of Elmer's glue to 100% acrylic paint would duplicate the old formula? I think the old ingredient was polyvinyl acetate.

I'm wondering if adding 1 ounce of Elmer's glue to 100% acrylic paint would duplicate the old formula?

No.

Paints are a specific mixture of a number of components designed to fulfill their role well in this specific combination. If you go and add a random additional ingredient, which in itself is not a pure substance but a nondisclosed mixture, you cannot predict what will happen. At best, the additional ingredient does no good and no harm. At worst, you can throw away the entire can of paint.

Paint companies also do not disclose key data about their mixtures for the obvious reason that they don’t want to be copied. So even though it may say X, Y and Z on the back of the can, that doesn’t mean you can infer much about the mixture. Even if they give percentages, you don’t know what all the rest is. Therefore, you cannot duplicate the old formula unless you have access to the full blow of analytical methods and a sample of the old paint.

Finally, I’m not even sure if changing vinyl acrylic on the packaging to just acrylic has any implication on the content of the can. It could just be a new choice of description.

Paint formulations continue to evolve to humor the EPA. Basically is is against EPA rules to make older more effective paints. Long ago I had some chlorinated rubber paint , it was great for rusty steel ; been illegal for decades.


Why does zinc react while its outermost suborbital is full? Why does zinc lose electrons while its outermost configuration is 4$\rm{s}^2$ 3$\rm{d}^{10}$ both outermost suborbital are full.

Why does it react or lose electrons?

You could ask the same question about the alkali earth metals with their $\rm{s}^2$ valence shell. While there is an energy cost to removing electrons, in a spontaneous reaction (e.g. $\ce{Cu^2+_{(aq)} + Zn_{(s)} -> Cu_{(s)} + Zn^2+_{(aq)}}$) there will be a compensatory transfer of energy elsewhere (like the reduction and deposition of $\ce{Cu^2+}$) which leads to a more stable overall system.

Zinc has quite a stable configuration, but that does not mean it does not want to attain noble gas configuration. You should also observe that $Zn^{++}$ is more stable that ground state of Zn. This is due to the fact that both $e^-$ from $4s^2$ orbital are removed and the Zn ion has a fully-filled d-orbital, which is much more stable than the ground state configuration.  Thus Zn reacts to get more stability provided by the fully-filled d-orbital.


What determines the high degree of selectivity in free radical brominations? In my organic chemistry class, we were asked what determines the relatively high degree of selectivity in the reaction of a bromine radical with primary, secondary, and tertiary C–H bonds.

I said it is the overall free energy of the reaction because according to Hammond's postulate, endothermic (endergonic) reactions have transition states closely resembling the product side, making the difference between primary, secondary, and tertiary radicals more important in the transition state.

Would this logic be wrong?

First, let’s take a look at the radical chain reaction that leads to the substitution in question:

$$\begin{align}\ce{Br^. + H-R &-> HBr + R^.}\tag{1}\\[0.4em]
\ce{R^. + Br-Br &-> R-Br + Br^.}\tag{2}\end{align}$$

Obviously, the selectivity is determined in the first step as long as no radical recombinations take place. If we break down the process in the first step into sub-processes, we get:

cleavage of the $\ce{C-H}$ bond;
formation of the $\ce{H-Br}$ bond.

Again, the second is the same no matter which radical is formed; therefore, it must be the cleavage of the $\ce{C-H}$ bond that leads to selectivity. Indeed, you can imagine radicals as being electron-deficient compounds and therefore behave similar to cations. Being similar to cations means that you can assume hyperconjugation to be a viable stabilisation mechanism for radicals; therefore, the energy lost by homolytically cleaving a tertiary $\ce{C-H}$ bond is slightly less (a less unstable product) than that lost cleaving a primary $\ce{C-H}$ bond. Exploiting this energy difference is the basis of selectivity.

This selectivity is higher if bromine is used than if chlorine is used. The reason is precisely the Hammond postulate you cited: the reaction with bromine is predicted to be more endothermic and thus the transition state more product-sided. A layman’s explanation would be that bromine can approach slowly and try out whether it feels happy about reacting with this hydrogen — think of playing Jenga and carefully testing if a brick is moveable or not. Chlorine on the other hand, is faster to react and thus has a less product-sided transition state.

There is a second factor that reduces selectivity. Assume 2-methylbutane; this compound conveniently has primary, secondary and tertiary hydrogens. Probabilities of bromine radicals to react with each of these hydrogens have been calculated; since I don’t have actual numbers on me right now, I am going to ballpark the following: assume, that a secondary hydrogen has a five times as high probability of reacting with bromine when compared to a primary one and that the tertiary hydrogen has a 15 times as high probability. Naïvely, we could assume that the reaction yields products with a ratio of $1:5:15$ — but we need to remember how many hydrogens of a given type there is.

The possible products are 1-bromo-3-methylbutane, 1-bromo-2-methylbutane (both primary attacks), 2-bromo-3-methylbutane (secondary attack) and 2-bromo-2-methylbutane (tertiary attack). For each of these, the following number of hydrogens can react to form the product in question:

1-bromo-3-methylbutane: 3
1-bromo-2-methylbutane: 6
2-bromo-3-methylbutane: 2
2-bromo-2-methylbutane: 1

Thus, to accurately calculate the ratios, we need to multiply the preference factor with the number of hydrogens, resulting in a product ratio of $3:6:10:15$ in the same order as presented — a lower fraction of tertiary bromide than presumed.

Mechanism proceeds by radical abstraction of hydrogen. The resulting carbon radicals follow carbocation stability ie 3>2>1. That being said, radical halogenation does not give very selective products, almost always a mixture. For example halogenating pentane will not give only 1 and 1,5 di bromo.


When are the terms diphosphate and pyrophosphate used? On the Wikipedia article about pyrophosphate, it states that:

  The group (pyrophosphate) is also called diphosphate or dipolyphosphate.

But for example, why is ADP apparently called adenosine diphosphate but not adenosine pyrophosphate? The same for geranyl pyrophosphate (GPP) instead of geranyl diphosphate?  In which context the terms diphosphate and pyrophosphate are used?

Great question, but unfortunately without a great answer.  Chemistry is very inconsistent in its naming of phosphates.  In addition to the terms diphosphate and pyrophosphate, the term bisphosphate is also used.

Someone will probably put in an answer that cites "official" IUPAC or IUBMB rules.  In many cases (but not all) these rules are more sensical and well thought-out than the terms in common usage in the field, but unfortunately, they are not in universal (or even wide) use.

Here is a rough breakdown of my sense of how these three terms are used:

Inorganic chemistry

Pyrophosphate indicates a phosphate anhydride.
Diphosphate indicates a salt that contains more than one (ortho)-phosphate anion.
Bisphosphate is not widely used.

Biochemistry

Pyrophosphate: used inconsistently to name an organic derivative of a phosphate anhydride.  This is what you see in geranyl pyrophosphate (GPP).  It is also used to refer to "inorganic" pyrophosphate anion, often abbreviated as PPi.
Diphosphate: an unfortunate term that is ambiguous, although usage is shifting towards synonymity with pyrophosphate.  It is often used to mean "pyrophosphate" and but sometimes still used to mean "bisphosphate".  Biochemically, the most important phosphate anhydrides are undoubtedly ATP, ADP, etc.  These are never referred to as pyrophosphates even though that term would be more accurate.  History has chosen "triphosphate" and "diphosphate" as the terms to use for adenosine-substituted phosphate anhydrides, icky though the term may be.
Older literature will also use "diphosphate" to mean "bisphosphate".  The name "fructose diphosphate was formerly in wide usage for what is now known as "fructose-1,6-bisphosphate", for example.  That compound contains no phosphate anhydrides.
Geranyl diphosphate is still a fairly widely used synonym for geranyl pyrophosphate.  You see abbreviations for this molecule of both GPP and GDP.  I prefer the name GPP, but unfortunately not everyone uses it.
bisphosphate: Now in wide usage as the preferred term for an organic molecule with multiple ortho-phosphate esters, such as fructose 1,6-bisphosphate.  Biochemists have done a relatively good job of adopting this usage in modern times.

A common confusion that stems from this nomenclature concerns the stability of organophosphate compounds. Phosphate esters, e.g. glucose-1-phosphate or glucose-6-phosphate or adenosine monophosphate, are much more stable and less reactive than phosphate anhydrides such as acetyl phosphate (a mixed anhydride of acetic acid and phosphoric acid) or the phosphoanhydride bonds in adenosine triphosphate (ATP).

Pyro- is for when both phosphate are together, di- means the phosphates can occur anywhere in the molecule.


According to this equation for glycolysis, what is reduced and where do the electrons go? I know that the glucose is oxidized into pyruvate but I don't understand what is reduced or where the electrons go....

Equation:
$\ce{Glucose + 2 ADP + 2 NAD^+ -> 2 Pyruvate + 2 ATP + 2 NADH}$

NAD+ accepts a hydride ion (proton + 2 electrons) and is reduced to NADH.

I think the question "where do the electrons go?" is excellent, and if answered correctly can excite otherwise disinterested students into chemistry. This is because the only way to properly answer the question is to get into the mechanistic details which are at the heart of chemistry. And it's not that hard, given a quality image to point out the specific "electron pair". I've included one of many images that can be found on the web for the mechanism of Glyceraldehyde-3-phosphate (G3P) Dehydrogenase below. The precise answer to the question is that the "electrons come from the bond attaching the hydrogen directly to the carbonyl carbon of G3P.  This corresponds to the same hydrogen in the open chain form of glucose, and that, in turn can spark further questions about that. Unfortunately, despite the central importance of this step, many (I would say most) educators would have trouble answering with this precision. No wonder chemistry gets a bad reputation.


Reaction of silver nitrate and hydrochloric acid Consider the reaction of $\ce{AgNO3}$ and $\ce{HCl}$. I read that silver chloride would be formed. But $\ce{H} > \ce{Ag}$ in reactivity, then how could $\ce{Ag}$ displace $\ce{H}$ from $\ce{HCl}$ ?

If there was a reaction, $\ce{AgCl}$ and $\ce{HNO3}$ were to form. But then these would react again to give us the original compounds back. The actual question was to write down the reaction between $\ce{AgNO3}$ and $\ce{HCl}$. I (wrongly) realized that the compounds wouldn't react because of the reasons stated.

One of the most difficult parts of chemistry is learning to recognize a type of reaction based solely on its reactants. This has to be done before you can apply a reaction pattern to the problem, and so it is critical to get this step right in the beginning.

In this case, you are looking at the reaction:

$$
\ce{AgNO3 + HCl ->  AgCl + HNO3}
$$

You already know the products, but are questioning how these products can be formed since the activity of H is greater than Ag, implying that Ag is easier to oxidize than H. This is true - but take a look at the oxidation numbers of each species in this reaction:

$\ce{Ag: +1 -> +1}$

$\ce{NO3: -1 -> -1}$

$\ce{H: +1 -> +1}$

$\ce{Cl: -1 -> -1}$

Since there is no overall transfer of electrons, this can't be a redox reaction, and that means activities won't matter here.

That means there must be another driving force for this reaction - another reaction pattern that fits better.

I don't want to give you the answer directly, but I will give you some advice that should help:

Write the full equation - including the phases.

$$
\ce{AgNO3(aq) + HCl(aq) ->  AgCl(s) + HNO3(aq)}
$$

See if you can find another reaction pattern that fits this equation better. Later, I'll update this answer to show how you can identify the correct pattern for aqueous reactions using only the reactants.

Update

As I mentioned, one of the hardest parts of chemistry is learning to identify reaction patterns based only on the reactants. This is a double-displacement (or metathesis, or precipitation) reaction. It is easy to see now, given that the reactants are aqueous and at least one of the products is solid (the precipitate). Formation of the solid is the driving force for this reaction - the quick explanation is that the forces attracting silver and chloride ions together are stronger than the solvation forces between those ions and water, as well as the forces holding them to sodium and nitrate ions.

This is good, but how do we predict that this is a precipitation reaction ahead of time?

The key is to have a good understanding of the three common types of salt/acid/base reactions that occur in aqueous solution, and to learn to recognize clues in the reactants.

Briefly, the three types that are most commonly seen are:

Precipitation
Acid/base (Arrhenius definition)
Redox (single displacement)

You can recognize each of these by the reactants if you know what to look for.

Precipitation - Two soluble salts (ionic compounds), or a salt with an acid or base.

Acid/base - An Arrhenius acid and base (compound containing hydroxide)

Redox - a salt or acid and an elemental metal

There are many other types of reactions that occur in aqueous solution, and many variations of the acid/base and redox category, but these three cover the cases most commonly seen in a classroom.

Once you have identified the likely pattern that the reaction will follow, the next step is to predict the products using that pattern to see if they make sense. If they do, then you have likely chosen correctly.

You can use this algorithm for more advanced chemistry as well - in organic chemistry, for example, one of the major goals is to learn to predict reactions based on functional groups. Once you can identify functional groups and have memorized reaction patterns for them, it becomes possible to predict a huge range of reactions.

The silver displaces the hydrogen because this is a double replacement equation, they just switch their anion. When the silver would not be able to displace the hydrogen is when It would have been a single replacement equation like Ag + HCl


Is it possible to freeze water by dissolving a salt? Theoretically, by dissolving a salt in water the melting point lowers, approximately 1.86 K kg/mol, making it more difficult to freeze water. However, the process of dissolution of certain salts is endothermic, lowing the temperature of the water. Is it possible that the water freezes due to that change of temperature although the melting point also lowers, or is it impossible?

For instance, consider the next case. In 1 kg of liquid water at 0 °C, we put a mol of KI. The melting point of water should be now about −1.86 °C. However, the enthalpy of dissolution of the KI is about 20 kJ/mol, meaning that the temperature of water should lower till −4.8 °C. So, theoretically, it should  freeze, but does this happen in reality?

It is possible, in a significantly different way that you envision. Freezing point depression via the cryoscopic constant is an example of a colligative property, which holds only for relatively dilute solutions. Once the solutions get very concentrated, intermolecular interactions become more complicated and do not generalize, meaning different compounds will affect the solvent differently.

Therefore, with enough of the right solute, it is possible to actually increase the freezing point of water. Oscar Lanzi's example of clathrates is interesting, but if you want to use salts, then I present you tetrabutylammonium hydroxide. It is known to form stable hydrates with a defined composition and containing a large amount of water, such as $\ce{(C4H9)4N+OH^-.30 H2O}$. The 30-hydrate is a solid which melts at approximately 30 °C, containing 67.6% water by weight.

This specific compound, at this specific concentration, forms a particularly stable network of solvating water molecules. You can think of it as stabilizing the structure of ice, allowing it to occur above the normal melting point of pure water. Thus, in principle, if you add enough of the anhydrous salt to water, when you reach the right ratio, the solution will "freeze".

Many compounds actually display this behaviour of forming solid hydrates at certain defined compositions, but tetrabutylammonium hydroxide is unusual in that it forms solid hydrates with a very large amount of water.

I don't think it is possible. My reason being that the world has many oceans and seas even salt water lakes. All of which vary in different densities and different temperatures but some form ice packs, ice burgs, glaciers yet others don't freeze at all. Again it's down to the type of salt density & composite of that salt surely???


Can an SN1 reaction occur in compounds with the leaving group attached to a primary carbon, which can later rearrange? Can SN1 reactions occur in compounds where the carbocation, even though is is initially $1^\circ$, can rearrange to form a more stable carbocation?

Examples cases:

Your thinking is correct that they can rearrange and then form stable carbocation (though your first example is wrong as it would become anti-aromatic after rearrangement) ,
but the answer to your question is NO , they would be substituted by SN2 mechanism being
1 degree . The reason behind this is that the compound will not try to form a carbocation first where it is not that stable , you can think of it in this way that at the initial stage the compound doesn't know that it can rearrange later to form stable carbocation, this is something that we can see but not the compound and thus it will go by normal SN2 mechanism.


Unusual Result in Kinetics Lab The question: In the lab described below, why did the decolorization reaction described, which should be first order in each of the two reactants, go most slowly when the reactants were most concentrated?

As part of the introduction to our kinetics unit, I asked my students for factors that they thought would influence the amount of time it takes a reaction to occur.  Not surprisingly, they came up with a variety of things, including temperature and the concentration of reactants.  I set up a simple lab to compare temperature and concentration conditions based on the reaction of crystal violet with sodium hydroxide.  Later in the unit we will come back to the reaction and analyze its kinetics more carefully, but for the moment, I just wanted my students to do a more basic analysis.

The students did trials for 4 conditions, and each involved the same quantity of reactants: 25 ml of 0.10 mM crystal violet and 25 ml of 1.0M sodium hydroxide.  The reaction was performed by adding the crystal violet to a flask, adding water to dilute, adding the sodium hydroxide to the flask, stirring, and timing the reaction until it no longer showed visible purple.  (Yes, that's a sloppy endpoint, but it was sufficient for this activity's purposes.)

Trial 1: 25 ml crystal violet, 150 ml room temp distilled water, 25 ml sodium hydroxide (a 1:8 dilution of stock reactants)

Trial 2: 25 ml crystal violet, 150 ml hot distilled water, 25 ml sodium hydroxide

Trial 3: 25 ml crystal violet, 50 ml room temp distilled water, 25 ml sodium hydroxide (a 1:4 dilution of stock reactants)

Trial 4: 25 ml crystal violet, 0 ml distilled water, 25 ml sodium hydroxide (a 1:2 dilution of stock reactants)

The results:

The hot reaction went much faster than the cold.
The 1:4 dilution went faster than the 1:8 dilution.  (~6 minutes vs 9 minutes, respectively)
The 1:2 dilution went much, much more slowly than any other. (~16 minutes)

Result #3 is the anomaly.  It involved the most concentrated solutions and so should have been the fastest of the 3 room temperature reactions, but it was far and away the slowest.

Things I've considered:

The reaction isn't significantly endo- or exothermic, so the solution
didn't have a different temperature due to its minimal mass.  (All of
the reactions except for the one with hot water added were within a
couple of degrees of each other.)
The students didn't make a mistake in execution - this was observed by multiple groups over multiple trials.
The difference was not due to the sloppy endpoint of the reaction.  Though the reduced volume meant that a given amount of remaining crystal violet would be most purple in this trial, the difference in color was enormous and obvious.
The reverse reaction never becomes significant with such a massive excess of sodium hydroxide.
New idea: The unusual result is the only trial without distilled water - maybe the water is contaminated or something like that, but in a way that is promoting the reaction?  I was excited when I thought of this (I actually popped up out of a dead sleep with it in my head,) but it seems unlikely, I used the exact same container of water an hour earlier to make each of the stock solutions...

I've wracked my brain, but I can't figure out a reasonable explanation for what we observed.  Any suggestions would be welcome.

Your analysis of the expectation is correct, and we can look at it more quantitatively in this way:
Since the hydroxide concentration is much higher than the CV concentration in all of the reactions, let's approximate it as unchanging. Assuming the reactions are first-order in each reactant, we have a rate law for the concentrated reaction of:

$$-\frac{d[\ce{CV}]}{dt}=k[\ce{HO-}][\ce{CV}]\approx k[\ce{HO-}]_{initial}[\ce{CV}]$$

For simplicity, we'll define a $k'$ as $k'=k[\ce{HO-}]_{initial}$ and treat the reaction as pseudo-first-order.

The half-life of a first-order reaction is given by $t_{1/2}\approx \frac{0.7}{k}$, so for the concentrated reaction we have $t_{1/2}=\frac{0.7}{k'}$.

For the 1:4 reaction, the initial hydroxide concentration is halved, so we have a pseudo-first-order rate constant of $0.5k'$ and $t_{1/2}=\frac{0.7}{0.5k'}$ or 2 times the half-life of the concentrated reaction.

Likewise, the half-life of the 1:8 reaction is twice that of the 1:4 reaction and four times that of the 1:2 reaction, so it's $t_{1/2}=\frac{0.7}{0.25k'}$.

Let's say that the concentrated reaction needs 6 half-lives for the concentration to get to the point where the color disappears. Since the 1:4 reaction starts at half the initial CV concentration, it only needs 5 half-lives. But the half-lives are each twice as long, so it should take a time equivalent to 10 half-lives of the concentrated reaction, so 10/6=1.7 times as long.

For the 1:8 reaction, the same analysis says that it will take 4 half-lives, but each is 4 times as long as that of the 1:2 reaction, so it takes 16 equivalent half lives, meaning 16/6=2.7 times as long.

The only instance in which the more dilute reaction will be faster is if the concentrated reaction requires less than two half-lives to be observed to be complete. That can't be the case here or there would be no color at the start of the 1:8 reaction.

So we must look for an alternative explanation for your observations. 0.5 M NaOH is a very concentrated solution, and it is possible that other reactions are occurring besides the intended one. To test that, I would follow Maurice's suggestion and run the reaction at a wider range of base concentrations. It would also be preferable to have the same concentration of CV in every reaction, so only one variable is involved. To do this, used a fixed volume of (NaOH + water) in each reaction.

UPDATE: This paper1 reports doing this reaction at much lower hydroxide ion concentrations and also notes that the reaction rate decreases with increasing ionic strength, which could be at least a partial explanation for why your observed rate at high concentrations is lower than expected.

1 Felix, L.D. Kinetic Study of the Discoloration of Crystal Violet Dye in Sodium Hydroxide Medium. J Chem Appl Chem Eng Vol: 2 Issue: 1

All the reaction have different concerntration at start . As you have taken so much excess NaOH it would remain constant. We could look it as you have a solution of 50 ml with 0.05 mM violet red and 0.5M NaOH . You dilute the N times and then to have to compare rates. Threotically we could find that but when we say discolouration in the solution does it mean the reaction has been 100% complete? Actually there is a particular concerntration at which we will see discolouration . Lets say that concerntration is 0.001mM of violet red.

Time taken in completion of reaction (assuming the reaction is 1st order with respect to violet red and also note that pseudo rate constant depends on intial concertration of NaOH)= $\frac{ln(\frac{intial\,contretaion}{final\,concentration})}{k(concentration\,of\,NaOH)}$ = $\frac{ln(\frac{0.05*1000}{n})}{(k*0.5/n)}$=$\frac{2n*ln(\frac{50}{n})}{k}$=F(n)

Here k is the kinetic rate of the 2nd order reaction.

F'(n) =  2ln($\frac{50}{n}$) -2

As you can the derivative of F is greater than zero till 50/e and after that less than zero
(You could assume the complete reaction concerntration be anything but assume it be constant)

So whatever concentration you choose the function comes out that time of completion of reaction increases(F'(n) may come out be negative at the very start though) till a particular value of n and after that decrease as n increase(n is directly proportional to dilution)

If we assume the kinetic 2nd order law to be completely valid we can find the concerntration of violet red at which our eyes think the reaction has been completed.Yeah for that you will need to conduct many experiments.

The result is not unusual and its perfectly fine. If you find the rate of the reaction at intial points I m sure you will find it to be greatest in most concerntrated solution.


Why do strong acids have weak conjugate bases? This question is based on the Brønsted-Lowry model. I saw many answers to similar questions online. However none of them have given a detailed answer that is based on molecular structure (the answers all told the OP to rely on a reference table for a way to identify strong and weak acids).

All chemical processes are reversible at least to some extent.

Keeping this in mind, a strong acid must have a weak conjugate base (I think you mean strong acids have weak conjugate bases).

Why? Let's try to prove by contradiction. If a strong acid had a strong conjugate base, then the base would quickly re-associate itself with a hydrogen proton. This reverses what the acid just did - dissociate itself from a hydrogen proton!

To illustrate this concept, consider hydrochloric acid - the oft-cited strong acid. $\ce{HCl}$ is a strong acid because in water solution it dissociates (nearly) completely into chloride ion and hydrogen protons (which are then solvated by water).

$\ce{HCl + H_2O ->H_3O^+ + Cl^-}$

That is the reaction of hydrochloric acid with water. Note the one-way arrow; the reaction is a one way street. $\ce{HCl}$ is the acid; $\ce{Cl^-}$ is the conjugate base; water is the base and hydronium ion is the conjugate acid. In other words, this is the generalized reaction:

$\ce{(strong)~acid + base -> conjugate ~acid + conjugate ~base}$

In bite-size pieces:

$\ce{(strong)~acid -> conjugate ~base}$

and

$\ce{base -> conjugate ~acid }$

Add the above two "half-reactions" together and you get a complete Brønsted acid/base reaction.

So back to the dissociation of $\ce{HCl}$.

If this reaction (the reverse reaction) happened to a significant extent:

$\ce{HCl + H_2O <- H_3O^+ + Cl^-}$

Then that would be antagonistic to the dissociation of $\ce{HCl}$! $\ce{HCl}$ would no longer be nearly completely deprotonated in water solution since the reverse reaction is significant!

Therefore, we conclude that if one has a strong acid, then its conjugate base must be weak. The converse is also true; if an acid's conjugate base is weak, then the acid must be strong.

Strong acids readily give up their protons, because the resulting conjugate base becomes more stable with respect to its new electron distribution. An example of this would be the acid halides with the exception of HF. The conjugate bases of these acids achieve their octets and diffuse polarizability, rendering them electronically stable.


Which is more stable Ethyl carbocation or Neopentyl carbocation? As per my knowledge, hyperconjugation is more dominating over inductive effect, so due to three alpha hydrogen ethyl carbocation should be more stable. In some books, while solving questions I observed that answers can match if I take neopentyl carbocation as more stable.

so which one of them is comparatively more stable?

In that case ethyl carbocation has two alpha hydrogen atoms as CH2 is bonded to C+.And in case of neopentynyl carbocation CH3 is attached to C+ .more alpha hydrogen atoms ,more hyperconjugative structures,and more stability.and there are three donating groups in neopentynyl carbocation.these may be the reasons


How to determine the order of acidity of the following dimethyl nitrophenols This is how I went about it, although I got a wrong answer:

3,5-Dimethyl-4-nitrophenol (2) shows the nitro group to be sterically hindered, due to which it should go out of plane and this will be an example of the steric inhibition of resonance. Hence, there will only be a weak withdrawing inductive effect, and no resonance. Thus, this should be the least acidic.
4,5-Dimethyl-2-nitrophenol (3) shows a hydrogen bonding between the nitro and the hydroxyl group, due to which there will be a lower tendency to release the proton. Hence, even this should be less acidic.
Between 2,6-dimethyl-3-nitrophenol (1) and 2,4-dimethyl-5-nitrophenol (4), (1) should be weakly acidic due to the hyperconjugation destabilisation caused by the two methyl groups, and hence the answer should be (4), with lower hyperconjugation, lower induction from the methyl groups, and higher induction from the nitro group (no resonance at the meta position).

However, the answer is given to be (2), which I ruled out first.

Can someone please explain what is happening, and where I've gone wrong?

The right answer should be 3 as the $\ce{NO2}$ group has a negative mesomeric effect, which is only possible with 3. In 2, due to steric inhibition of resonance, it exhibits only -I effect, and although 3 contains H-bonding, it slightly affects acidity.

$\ce{-NO2}$ is an electron-withdrawing group and $\ce{-CH3}$ is an electron-donating group and the effect of these groups can be observed at ortho- and para- position of the ring with respect to the functional group attached to the ring.

in the above resonating structure of toluene you can see -ve charge in o-, p- position, similarly in nitrobenzene you can see positive charge in o-, p- positions, and $\ce{-OH}$ group attached to electron deficient carbon or positively charged carbon release $\ce{H+}$ more easily. Considering all this condition second option is the only option where carbon to which $\ce{-OH}$ is attached is getting positive charge and not getting a negative charge, so $\ce{-OH}$ is ready to loose $\ce{H+}$.


Identifying coordinate bonds How can we know whether a compound has a coordinate bond or not? I know that I have to draw the Lewis Structure for the compound. But what to do after that? Is there anything to do with formal charges?

The first thing needed for a dative bond is a lone electron pair in one atom and an empty orbital in another atom.
This condition is necessary but enough.

There is also a formula that fails in many cases but may sometimes become handy:
Formal Charge = (Oxidiation State of the centeral atom - 8 + Digit of its group number)/2

e.g.: For SO3 it would be: (6-2)/2 Which means that it has 2 co-ordinate bonds.
However for a compound like NH4+ it won't work.

Of course, safest way is to find out how the compound had formed in the first place.
e.g.: If we know that BF3 + F- made BF4-, we would find out that F- had shared with B two of its electrons, filling the empty orbital belonging to B.

After drawing the structure, you can basically check if the shared pair of electrons are coming from the same molecule. If the answer is yes, then the compound contains coordinate covalent bond.


Does the potential energy increase when temperature is raised? The potential energy of a material is its ability to do work. So I would suspect the potential energy to increase during phase changes that increase entropy (solid $\rightarrow$ liquid and liquid $\rightarrow$ gas), and decrease in the opposite direction.

However, what about when the temperature is raised for a solid, up until the melting point - will the potential energy increase? The heat from a 1500 K gold plate could do work on liquid water (evaporate) to a higher extent than a plate at 300 K, so I guess the potential energy has increased. So would the potential energy as a function of temperature, of an arbitrary material, look something like this ($K$ should be $T$)?

Yes, potential energy increases with increasing temperature for at least the following three reasons:

At a higher temperature, more atoms/molecules are in excited electronic states.  Higher electronic states correspond to greater potential energy. Potential Energy is -2 times Kinetic Energy. So actually, at higher temperature, when more atoms are in higher electronic states, there is more potential energy and less kinetic energy (just considering electronic energy).
At higher temperature, more molecules are in excited vibrational states.  Higher vibrational states correspond to greater potential energy. Half the energy of each vibrational state is kinetic, half potential.
At higher temperature, more molecules are in excited rotational states.  While for a rigid rotor, this would not involve increased potential energy, real molecules are not rigid.  The rotation of a diatomic molecule about an axis perpendicular to the bond stretches the bond as a centrifugal distortion, which represents an increase in potential energy.

Additionally, for intermolecular forces, to the extent that they involve potentials of the form r^x, the Virial Theorem requires that total energy is divided into potential and kinetic energy in a specific proportion, so total energy can not be increased without increasing potential energy.

Free translational motion is not subject to a potential, nor is rigid rotation.

Lets talk about a heating curve
As heat is constantly added to a substance in the solid phase, the KE increases but the PE remains the same. As the substance reaches its melting point, commonly known as fusion (border between melting and freezing), the KE remains the same while the PE increases. After a while the substance becomes completely a liquid, and heat is still being added constantly and the KE increases while the PE remains the same. The substance now reaches its boiling point, and starts to change into the gas phase, in this process the KE remains the same while the PE increases.

KE = kinetic energy
PE = potential energy


Werner's theory and complex salts According to Werner's theory, metal ions have two types of valency-primary and secondary, where the primary valency is said to be satisfied by negative ions only, and  the secondary valency can be satisfied by positive ligand, negative or neutral molecule.

My question here is:
While formation of complex compounds, which of these valencies is satisfied?

What I'm not able to interpret from this is whether there is a 'possibility' for both the valencies being satisfied or is it necessary that both are satisfied?

Also if the later is true how is it possible? Can it be explained taking some example as reference.

I think I'm not able to interpret it and understand its details.

A certain number of ions, atoms or molecules closely associate around a central atom leading to the formation of distinct entity called coordination complex.
The groups atoms or molecules linked to the central atom are said to be coordinated with the latter and their number gives the coordination number.

The non-ionisable units lie in the primary zone or sphere, whereas the  ionisable  units lie in the secondary sphere and appear as ions in the solution.

In hexammine platinic chloride $\ce{[Pt(NH3)6]Cl4}$  and in potassium ferrocyanide $\ce{K4[Fe(CN)6]}$ four chlorine ions of the former and four potassium ions of the latter ionize in the aqueous solution, whereas six ammonia molecules of the former and six cyanogen groups of the latter belong to the primary sphere or non-ionizable sphere and thus fail to respond to their characteristic tests or fail to ionize. Or you could also say that the chlorine ions in the former and the potassium ions in the latter occupy the primary valencies and hence are ionizable and the ammonia and cyanogen group occupy the secondary valencies  and are non-ionizable, while the $\ce{Fe}$ atom in the latter is present in the primary Sphere and thus fails to ionize.

As per your question during the formation of a compound say $\ce{[Pt(NH3)5Cl]Cl3}$, five molecules of ammonia and the chlorine ion (linked by secondary valencies) are present in the primary Sphere and the rest of the three chlorine atoms lie in the secondary sphere. Thus the coordination number of the compound would be  six as five ammonia molecules and one chlorine atom are attached to the central metal atom.

Now, every element tends to satisfy both its primary and secondary valencies. A negative ion tends to satisfy both these valencies i.e. primary as well as the secondary valencies.
The presence of a negative ion in the coordination sphere reduces the amount of charge on the complex ion by the amount of charge present on it while the negative ions in the primary sphere or the coordination sphere are not ionized.

Werner's theory proposed by alfred werner  in 1898.He was awarded by noble prize in 1913. It explains about the formation of coordinate complex compounds, bonding of coordinate complex compounds, stability of coordinate complex compounds and isomerism of coordinate complex compounds.

According to Werners theory each coordinate complex compound exhibits two types of valencies.The are a) primary valency and b) secondary valency.
 Primary valencies are satisfied by only negative charge and it is equal to oxidation state of central metal ion.

Secondary valencies are satisfied by either negative charge or neutral species and it is equal to the coordinate number.
Primary valencies are denoted by dotted lines and ionisible.
Secondary valenices are denoted by dark or straight lines and non ionisible.

In coordinate complex compound formed  bonds are coordinate covalent bonds.
In coordinate complex the central metal atoms acts as the Lewis acid and the ligand acts as the Lewis base.
If coordinate complex shows six ligand in it then the structure is octahydral and four ligand in it then the structure is tetrahydral.


Questions on concentration Currently I am attempting two questions on concentration but I am not sure if I am correct. Can someone help me with these questions? These are the questions:

  1) Calculate the volume of pure water added to dilute $\pu{0.2 M}$ $\ce{HCl}$ solution in $\pu{250 cm3}$ to $\pu{0.12 M}$ $\ce{HCl}$ solution.

$\pu{0.2 M}$ solution contains $\pu{0.05 mol}$. $\pu{0.12 M}$ solution contains $\pu{0.03 mol}$. The difference between the two solutions is $\pu{0.02 mol}$ (Which is how much water was added), so

$$V(\ce{H2O}) = 0.02 \times (2 \times 1 + 1 \times 16) = \pu{0.36 L}$$

  2) Calculate the mass of $\ce{Na}$ that will form $\pu{0.15 M}$ solution of $\ce{NaOH}$ by reacting with $\pu{400 cm3}$ of water.

I know that this method is totally wrong but I'll show you:

$$\pu{0.15 M} = \frac{n}{0.4} \implies n = 0.06$$
$$m(\ce{Na}) = 0.06 \times 23 = \pu{1.38 g}$$

where $n$ just refers to the concentration ($C = n/V$).

Edit:

Thanks to @George Tian I have made another attempt for the second question.

I have calculated the number of moles of NaOH for a 0.15M solution:

c = n/v
n = 0.4 x 0.15
  = 0.06

And I calculated how much Na is needed for this reaction:

The balanced equation is: 2 Na + 2 H20 -> 2 NaOH + H2
So, 0.06 mol of Na is needed (Due to the molar ratio)

Then I converted moles to grams

Mass = 0.06 x 23
     = 1.38 g

To clarify, I've reworded your questions. Please comment if I changed the meaning:

Calculate the volume of pure water that needs to be added to dilute a  0.2M HCl solution with a volume of 250cm3 to a 0.12M HCl solution

The first step would be to work out how many moles of HCl is in the original solution. As you have stated

c = n / v

c = 0.2M

v = 250cm3 or 0.250L.

Therefore n = 0.05m

Then work out how much water is needed to make a 0.12M solution.

c = 0.12M

n = 0.05m

Therefore v = 0.42L

To work out how much water needs to be added, subtract the original 0.25L from 0.42L, which equals 0.17L

Calculate the mass of Na needed to react with 400cm3 of water to form a 0.15M NaOH solution.

As you haven't progressed much on this question, I'll give a few hints.

Write a balanced equation

Work out how many moles of NaOH is needed to make 0.15M solution

Work out how many moles of Na is needed to form that many moles of NaOH

Convert from moles to mass.

Hope this helps.

There is no way to take a 0.2M solution of anything and get a 0.12M solution. Although it is trivial to take a 0.20M or even a 0.200M solution and dilute it to 0.12M. Have you ever heard of significant digits?
You state that a 0.12M solution contains 0.03 moles...so if I get a million tons of 0.12M solution or a millionth of a microgram of 0.12M solution they both contain 0.03 moles of solute?? I. don't think. so. In a dilution the number of moles of the solute do not change (of course, I'm talking about a one pot dilution, in serial dilutions, you only use part of the solutions, so the amount of solute does change). If you start with .05 moles of solute, then you will end with 0.05 moles of solute (ignoring losses).
So, you calculate how much HCl you have (moles): 0.20 x 0.250 = 0.050 moles
Then you want to convert that to a concentration of 0.12 moles/L: 0.050/x = 0.12 or x = 0.05/0.12 = 416.67 mL (during calculations, you should bring thru more digits that you need in the result). You started with 250 which means you need 416.67-250.00 = 166.67 ml or 167 ml after adjusting for significance.
When solving these kinds of problems, I generally try to convert to things which are conserved. Atoms are conserved, density and concentration and volume are not, so converting from concentration to moles means you're less likely to make an obvious error. At least, that's my experience.


Steric Inhibition of Resonance in ortho-substituted Benzoic Acids It is given that due to Steric Inhibition of Resonance (SIR), the acidity of ortho-substituted benzoic acid increases with respect to benzoic acid. However, I'm unable to understand the reason behind this fact.

We know that resonance stabilises a molecule and the absence of which decreases the stability. In case of ortho substituted benzoic acids, resonance is inhibited due to the loss of planarity due to the steric interactions from the ortho substituent.

If this is the case then the conjugate base of the acid will not be stabilized by resonance with the benzene ring. So it seems the ortho-substituted benzoic acid will not readily donate H+ and is less acidic than benzoic acid, contrary to what is being given in various sources.

Then kindly explain why is ortho-substituted benzoic acid is more acidic than benzoic acid.

Important information before understanding the reason for steric inhibition of resonance and the ortho effect

In general, resonance stabilizes a molecule. But in the case of carboxylic acids, the conjugate base (anion after losing $\ce{H+}$) there are two equivalent resonance structures.
Take the example of acetic acid, $\ce{CH3COOH}$, and thus acetate ion, $\ce{CH3COO-}$:

The hybrid would look something like this:

It could be easily seen that equivalent resonance structures mostly stabilise the hybrid greatly.
Therefore, equivalent resonance tends to stabilize a molecule/compound more than normal resonance. (in most cases)

Now let's take the case of o-toluic acid.

There are two ways by which resonance can take place:

In (I), it takes place through $\ce{-COOH}$. Its ion is expected to be more stable due to the above result. Also, here $\ce{OH}$ is highly polar and it will tend to lose $\ce{H+}$ to become uncharged and thus be more stable (as compared to $\ce{O-}$).
In (II), it takes place through the ring. Its ion is less stable due to more formal charge and less aromatic character (due to + on the ring). Also, $\ce{OH}$ is less polar than $\ce{H+}$ due to the above point.

Therefore, it will be better to prevent the taking place of resonance through ring.

That is done by $\ce{-CH3}$ group which provides bulky group repulsion and takes the ring out of the plane of $\ce{-COOH}$ and thus providing condition for equivalent resonance in $\ce{-COOH}$.


Beer's law: UV-Vis absorbance values exceeding 2 In classical analytical chemistry, absorbance values in Beer's law can theoretically range from 0 to infinity.

As general rule of thumb neither high, nor very low absorbance are used for constructing calibration curves. The reason for avoiding high absorbance is that when absorbance is 2, only 1% light reaches the detector and with 3, only 0.1% light reaches the detector. Analytical chemists were taught to avoid absorbance > 1.5.  Low absorbance values are avoided because it hard to distinguish the light beam with and without the cuvet in place. The question then arises how much is the real absorbance and how much is contributed by noise.

Nowadays (= within the last decade), I am seeing UV-Vis spectrophotometers have absorbance ranges from 0 to 3, and some have even more. One day a PhD student showed an absorbance > 5 for a spectra, apparently without even thinking twice on the meaning of absorbance. Similarly, researchers showed an absorbance spectrum with the maximum at absorbance of 3.

My question is this a fundamental flaw in Beer's law that high absorbance values are erroneous or it is limitation in detector technology and the intensity of light sources? Please ignore the chemical reasons for deviations in Beer's law.

Imagine, if we had very high intensity light source, such as xenon arc lamp, instead of a typical deuterium or a tungsten lamp, would we get less error in the absorbance measurements of concentrated solutions? Absorbance is a ratio technique, but the number of photons reaching the detector will be high, even with conc. solutions, with intense light sources.

Can any spectroscopist shed some light on this issue (no pun intended) whether high intensity sources would beneficial in UV-Vis spectrophotometry of concentrated solutions?

Beer's law limitation can be more easily understood if the coefficient $\epsilon$ is considered as a cross-section effect. Let's start with a broad beam or a bundle of rays having a section $A$ (in $m^2$). Its measured intensity is $I_o$. Now suppose there exist plenty of tiny obstacles suspended in the bundle of rays, and that each such particle behave like a dark disk perpendicular to the bean, with a surface or a cross-section $\sigma$. $\sigma$/A is the fraction of $I_o$ that is blocked off by one such particule. If there are $n$ particules in the whole cylindrical bundle, the fraction of $I_o$ that is captured is $\Delta$$I$/$I_o$ = $n·\sigma$/$A$.

Now let l be the length of the bundle where the particules are present. The volume of this bundle filled with dark particles is $V=A·l$. So the fraction of $I_o$ absorbed by the particules is $\Delta$$I$/$I_o$ = $n·\sigma ·l/A·l$ = $(n/V)·\sigma ·l$ = $c·\sigma ·l$, if $c = n/V$ is the concentration of the particules in the volume crossed by the beam.

Going to the limit, this expression becomes : $dI/I_o$ = - $\sigma ·l·dc$.
Integration gives : $ln(I/I_o)$ = - $\sigma·c·l$

This last expression is similar to Beer's law :  $log(I/I_o) = - \epsilon ·c'·l$. I have written $c'$ because in the beginning of my text $c$ is the number of particules per volume unit. And the concentration in Beer's law is in mol/L. $c' = c/N_o$, where $N_o$ is Avogadro's number.

In the end we obtain the final formula : $\epsilon = N_o·\sigma /2.3$

Now if a particule is a disk of $0.2$ $nm$ diameter (dimension of a small molecule), the corresponding $\epsilon$ is about $10^5 L mol^{-1} cm^{-1}$. This is the order of magnitude of $\epsilon$ for the usual dye.

But if $\epsilon$ = $N_o·\sigma /2.3  = 10'400 m^2/mol$. And this is the point where I would like to go today. It means that if one mole of the dye molecules were all to be touched by the beam, they would have to cover a surface of $10'400 m^2$. This is as large as a football ground. It also means that when the concentration becomes too high, Beer's law does not apply any more. The majority of the molecules of the dye are hidden behind others.

You may even calculate that when $c > 1/ (\epsilon · cm), $ then $  : I/I_o < 0.01$, the absorbance is $> 2$, and this is the limit of application of Beer's law, because most of the molecules are hidden behind some other ones at higher values of $c$.

Sorry for the length of the demonstration.


Why is ice cubical in structure at very low temperature? My book says that at atmospheric pressure ice crystallises in the hexagonal form but at very low temperatures it condenses to cubic form.

Why and how does this happen?

First, a little clarification. The kind of ice you see at ambient pressure, ice I, exists as stacks of bilayers of water molecules. If they are stacked in an ABAB sequence, they form hexagonal ice, and cubic ice in an ABCABC sequence. The difference between the two allotropes is just the stacking of the molecules, so in fact they are structurally quite similar. The structure of cubic ice is analogous to the diamond structure of carbon, with hexagonal ice being analogous to lonsdalite. This picture is complicated by the fact that ice is generally proton-disordered, i.e. the molecules can have different orientations that break the crystal symmetry, subject to certain rules (the "ice rules").

There is evidence that ice crystallises as cubic ice in the upper atmosphere, namely the angle subtended by solar haloes and the angles formed in nascent snowflakes, for example. However, more recent evidence suggests that cubic ice doesn't not exist as a pure phase, but only in stacking-disordered hexagonal ice (http://pubs.rsc.org/en/content/articlelanding/2015/cp/c4cp02893g#!divAbstract). Curiously, there is also evidence that the lonsdalite form of carbon doesn't exist in reality, and is just stacking-faulted diamond (http://www.nature.com/articles/ncomms6447).

This is all very well, but does not tell us why hexagonal ice forms in preference under certain conditions, or cubic ice (?) under others. If you were to perform density functional theory calculation (essentially approximately solve the Schrodinger equation for a many atom system), you would find that hexagonal and cubic ices have almost identical energies, and are essentially degenerate at zero temperature, which makes it even more confusing why hexagonal ice is almost exclusively seen on Earth. I don't think there is a convincing answer to your question, since the best calculations do not show why the hexagonal structure of ice is lower in energy than cubic, but it could be due to a number of reasons: orientational disordering, which complicates modelling ices, zero point energy or free energy considerations, or a breakdown in the approximations used to perform such calculations. In any case, it looks like cubic ice doesn't exist under terrestrial conditions.

[PS. I do not know much but I'll still try to answer based on my logic and my prior knowledge]

The molecular structure of ice is Hexagonal always. At higher temperatures however, the molecule condenses to such as extent that it appears cubical. Refer to the picture

Temperature is the average kinetic energy of the molecules/particles in a compound/substance. At very low temperatures, the kinetic energy of the molecules tend towards 0. Also when a material freezes, the intermolecular bond length between 2 water molecules would decrease hence condensing the entire molecule.

Hope thats helpful.
More info at This site

If your question was about the shape of a bigger ice, I think that the shape of the ice also depends upon the container than you put the water in to freeze.


What is the correct way to pronounce iodine? I have always said it like 'io-deen' because whenever 'ine' is used (eg. amine, or fluorine) it is said that way. Sometimes I'm corrected by people who think it should be pronounced 'io-dyne', which is correct?

Manshu is correct,in British English we pronounce iodine as in seen while in North American English we pronounce it as in mine.

Think about how other halogens are pronounced: fluor-een, chlor-een, brom-een, astat-een, and newly named tenness-een. Why would it be io-dyne and not iod-een?


How to find number of moles of HNO3 behaving like acid in a redox reaction? In the reaction $(1)$, how to find the number of moles of $\ce{HNO3}$, which is behaving like acid?

The reaction is given as follows:
$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O} \tag1$$
To find number of moles of $\ce{HNO3}$ behaving like acid, I first balanced the reaction using oxidation number method which gave me:
$$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$
But I could not proceed any further. I am unable to find any hint. How should we proceed to reach the answer? Answer mentioned in book is 9.  This is a question from chapter "Redox Reactions".

It is always easy to balance redox equations by considering two half reactions. The equation to be balanced is:
$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O}$$
Thus, the two relevant half reactions to this redox equation are:
$$\ce{Sn (s) -> Sn^2+ (aq)} \\
\ce{NO3- (aq) -> NH4+ (aq)}$$
First balance elements other than $\ce{O}$ and $\ce{H}$. In this case they are $\ce{Sn}$ in first equation and $\ce{N}$ in second equation. Then, balance $\ce{O}$ and $\ce{H}$ in both equations with water and $\ce{H+}$, respectively. This is because the reaction is in aqueous acidic medium:
$$\ce{ Sn (s)  -> Sn^2+ (aq)}\\
\ce{NO3- (aq) + 10 H+ (aq) -> NH4+ (aq) + 3 H2O (l)}$$
Finally, balance the positive charges by electrons so that each equation has no net charges:
$$\ce{ Sn (s)  -> Sn^2+ (aq) + 2e-} \tag1$$
$$\ce{NO3- (aq) + 10 H+ (aq) + 8 e- -> NH4+ (aq) + 3 H2O (l)} \tag2$$
These are your mass and charge balanced oxidation ($(1)$) and reduction ($(2)$) half-reactions. Now, you can add these two equations in order to cancel electrons. To do so add $4 \times (1) + (2)$:
$$\ce{ 4 Sn (s) + NO3- (aq) + 10 H+ (aq) -> 4 Sn^2+ (aq) + NH4+ (aq) + 3 H2O (l)} \tag3$$
This is your balanced ionic equation. As evidence, you need nine $\ce{NO3-}$ ions to balance as counter ions in right hand side, so you need to add $\ce{9NO3-}$ to both side to keep the mass and charge balance.:
$$\ce{ 4 Sn + NO3- + 10 H+ + 9 NO3- -> 4 Sn^2+ + NH4+ + 9 NO3- + 3 H2O} \tag4$$
When simplify, the equation $(4)$ would look like:
$$\ce{ 4 Sn +10 H+ + 10 NO3- -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O} \tag5$$
Since $\ce{ +10 H+ + 10 NO3- = 10 HNO3}$, your complete balance equation is:
$$\ce{ 4 Sn +10 HNO3 -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O}\tag5$$
Therefore, your answer is correct. Answer in your textbook must be a misprint or accidently for got that one $\ce{HNO3}$ is needed to be reduced. Other $\ce{9NO3-}$ is to balance the products (as counter ions). All $\ce{10H+}$ ions needed to complete the redox reaction.

While balancing,
$$\ce{Sn + HNO_{3} \longrightarrow Sn(NO_{3})_{2} + NH_{4}NO_{3} + H_{2}O}$$
first balance oxidising agent and reducing agent only, we will get $\ce{4Sn}$ and $\ce{1HNO_{3}}$ , after balancing them, write number of extra moles of acid obtained while balancing other atoms. That will give us the answer.


How would this reaction look like in a potential energy diagram? (SN1 Reaction) My task is to explain, why molecule A reacts in a SN1-reaction faster than molecule B. (I would do this by comparing their potential energy diagrams)

My idea is to have a closer look at the reaction mechanism at first:

TS1 is here the intermediate with the carbo-cation. That's the step, where the reaction is slow, where one needs a lot of energy to remove the Bromine.
Maybe I guess one does not need this lot of energy in molecule A compared to B because the Bromine is in the background (trans- isomer) there and not in the foreground? Furthermore, in SN1 reactions the reaction rate is determined by the substrate (the first step).
So, the activation energy of molecule A is smaller then the activation energy needed in B and that's the reason, why the reaction ist faster?

An interesting question is how (or if) one really knows if both reactions proceed via bonafide SN1 type mechanism. Under the assumption that both reactions proceed via bonafide SN1 mechanism, we must infer that the carbocation (carbenium ion) intermediate 1 in both cases must be the same. While the previous answer correctly assumes that this intermediate may be described well as planar (at the carbon centre), the attack by water is the faster step and hence should not affect the overall rate of the reaction. Since both A and B must proceed through the same intermediate, (although not rate determining!), both must react at the same rate with the incoming nucleophile.

A better explanation can be deduced by looking at the most preferred conformations in the two starting materials. These are designated 2 and 3 below, for A and B respectively. In both cases, the bulky tertiary butyl group will likely prefer to be sat at an equatorial position to minimise 1,3 diaxial interactions. Once tertiary butyl group is "locked" at the equatorial position, we have to place the bromine axially (up) in A (2), and equatorially (down) in B (3).It is possible to argue (although still very qualitative and hand-wavy) that 3 is likely more stable than 2, since both groups are equatorial in 3.

Now, the rate of a reaction depends on the activation free energy barrier, which is the difference between the free energy of the reactant and that of the transition state. At this stage, without computation it is impossible to predict which transition state will be at a higher or lower energy. However, speculatively, to a first approximation the energies of the transition states would be very similar. Under this assumption, the activation energy barrier for A (higher energy conformation) should be less than that for B, which should explain the observed reaction rates.

Having said that, I would not necessarily think that energy diagrams are the best way to explain this difference in reactivity, or indeed if there is a single, simple explanation.

I would rather not solve this by comparing  potential energy values.

Instead there is another theoretical way of doing so.

When SN₁ happens, either the attacking reagent( In this case water) can attack from above or below the planar carbocation.

In the first case, Br⁻ leaves from above, and now water can easily attack from below.

In the second case, Br⁻ leaves from below, and water has to attack from above(to avoid steric repulsions between the exiting Br⁻ and attacking water), but now there are steric repulsions between C(Me)₃ group and the attacking water, leading to a reduced SN₁ rate as compared to the first case.