brightsky's Chem Thread

[lzxnl]

Like? If you have ethanoic acid, writing Cl-(eth) or something should indicate it's dissolved but not aqueous.
Or if you had ammonia, Cl-(amm) should be clear enough.

[brightsky]

What is the equation for the combustion of sulfur? I was always under the impression that sulfur was an octatomic molecule. So I wrote: S8(s) + 8O2(g) --> 8SO2(g). But the (VCAA-endorsed) solutions say: S(l) + O2(g) --> SO2(g). Which is correct?

[Aurelian]

What is the equation for the combustion of sulfur? I was always under the impression that sulfur was an octatomic molecule. So I wrote: S8(s) + 8O2(g) --> 8SO2(g). But the (VCAA-endorsed) solutions say: S(l) + O2(g) --> SO2(g). Which is correct?

I would say both are correct, and simply relate to the conditions under which each combustion occurs (the latter being at above the decomposition temperature of S₈ while the former being below it). I'd actually probably say that what you've written is preferable to the "VCAA-endorsed" answer, and I certainly wouldn't imagine you'd be marked wrong for it (unless there was some extra information in the stem of the question which made the liquid phase combustion more realistic/likely).

As an aside, although sulfur is usually found as S₈ in its natural, solid form, it can also be found in other forms too (e.g. S₇, S₆ and even S₁₈!)

[brightsky]

I would say both are correct, and simply relate to the conditions under which each combustion occurs (the latter being at above the decomposition temperature of S₈ while the former being below it). I'd actually probably say that what you've written is preferable to the "VCAA-endorsed" answer, and I certainly wouldn't imagine you'd be marked wrong for it (unless there was some extra information in the stem of the question which made the liquid phase combustion more realistic/likely).

As an aside, although sulfur is usually found as S₈ in its natural, solid form, it can also be found in other forms too (e.g. S₇, S₆ and even S₁₈!)

Ahh okay, thanks Aurelian!

I just flipped through the textbook, and found the chemical equation for the oxidation of sulfur during the production of sulfuric acid, which incidentally matches the one provided by VCAA. I guess at sufficiently high temperatures, the covalent bonds within S8 breaks, leaving just S?

Also, I'm beginning to think now that knowledge of the production of sulfuric acid is required...since I've come across several exams which ask us to explain the contact process, and give equations for the various steps involved. Thus far, I haven't come across any questions asking about the production of ammonia or of nitric acid, which seems a bit unfair, given that different schools select different chemicals to study in detail. Can someone confirm/confute this?

[brightsky]

Also, I have a question regarding the following question:

The most suitable analytical method of testing for the presence of ethanol in a sample of petrol is:
A. Gas chromatography
B. Paper chromatography
C. Flame test
D. Atomic absorption spectroscopy

Why is B wrong? I know ethanol is a volatile substance, hence making GLC a viable option, but ethanol is also polar, so why can't the presence of ethanol be tested using paper chromatography?

Thanks!

[lzxnl]

Also, I have a question regarding the following question:

The most suitable analytical method of testing for the presence of ethanol in a sample of petrol is:
A. Gas chromatography
B. Paper chromatography
C. Flame test
D. Atomic absorption spectroscopy

Why is B wrong? I know ethanol is a volatile substance, hence making GLC a viable option, but ethanol is also polar, so why can't the presence of ethanol be tested using paper chromatography?

Thanks!

Well...there are several complications with paper chromatography. Ethanol is colourless and you may well need UV light to see where the ethanol dot goes to. Also, petrol contains lots of different hydrocarbons and these may all be quite similar in chemical structure. Therefore paper chromatography may not be sensitive enough to separate the compounds.

[brightsky]

1. What are the state symbols of the species involved in the condensation reaction between glucose and fructose?
2. When asked to draw the structure of a particular alpha amino acid, should we draw the alpha amino acid in its zwitterionic form?

Thanks!

[psyxwar]

1. What are the state symbols of the species involved in the condensation reaction between glucose and fructose?

Shouldn't they both be aqueous? Both are water soluble, and in biological systems at least the process occurs in solution.

[lzxnl]

Agreed. For simple sugars, from a VCE perspective they're very soluble due to the number of polar OH groups that can hydrogen bond with water.

As for the structures of alpha amino acids, if they don't specify a pH, then draw it in the zwitterionic form. Generally the pKa of the first dissociation is around 2 and the pKa of the protonated amine is around 10. Therefore most of the amino acid molecules will have a COO- and a NH3+ group.

If there are acidic or basic sidechains, well you'd really need concrete figures for those. Generally a COOH sidegroup won't be as strong an acid as the first acid dissociation. However VCE oversimplifies and just says that in acid solution draw the COOH groups as COO- and in base solution draw the NH2 groups as NH3+, and put in both for neutral pH levels, which is really poor.

[brightsky]

Thanks!

Also:

There are two major types of proteins, fibrous and globular. Fibrous proteins are water-insoluble, long and stringy. Globular proteins are water-soluble. Explain why serine is more likely to be found in globular proteins than tyrosine.

[lzxnl]

Just look at the polarity of the side chains. Serine has a polar OH group attached to just one carbon, whereas tyrosine contains a large non-polar benzene ring, which is evidently more water-insoluble.

[Patches]

How do diprotic substances work in relation to pH?

For instance, if I have excess H2SO4, I understand it won't fully ionise to SO4^2- + 2H+

that is, all of the first H+ will be removed, but not all of the second - is that right?

How do we calculate the pH then?

Also, do bases have the same effect? For instance if I have excess Ca(OH)2, will it fully ionise to Ca2+ + 20H-

or will the second hydroxide ion not be released from all of the calcium hydroxide?



Hope that made sense

[lzxnl]

OK. Firstly, you'll have to know something about the relative acidities of the first, second and third (if any) dissociations of a polyprotic acid. For a weak acid like phosphoric acid, the first dissociation is already weak. Therefore the second and third dissociations will be so much weaker and their pH contributions can be ignored.

For sulfuric acid, however, HSO4- is actually a rather strong acid, around 1000 times stronger than ethanoic acid and has a pKa of pretty much 2. Therefore its dissociation cannot be neglected.

Let's say we have a 0.5 M solution of sulfuric acid. The first dissociation is strong; its pKa is -3, so we can just say that we have 0.5 M HSO4- and 0.5 M H+. Stuff the water autoionization. Water's pKa is 14.

Let the concentration of SO4 2- formed be x M. Then we form x M of H+ and so the final concentration of H+ is 0.5 + x.
By mole ratios, the final concentration of HSO4- is 0.5 - x
By the equilibrium law, 10^-pka = 0.01 = [H+][SO4 2-]/[HSO4-]
= (0.5+x)*x/(0.5-x)
So 0.005 - 0.01x = x^2+0.5x
Solving this equation yields x = 0.0096 M
Therefore the final concentration of H+ is 0.5096 M for a pH of around 0.3. Note how this value is different from the calculated pH of 0 that would arise if we assumed double dissociation.

As for bases, if you have a diprotic base, like say ethylene diamine (NH2-CH2-CH2-NH2), then a similar thing results; each NH2 group is basic but obviously the second protonation will be weaker as the positive NH3+-CH2-CH2-NH2 ion is trying to grab a positive H+. However calcium hydroxide is a bad example. Its only reaction with water is through dissociation, and there's a solubility product which, for a given temperature, states what the value of [Ca 2+][OH-]^2 is (equilibrium concentrations, note the squared term for hydroxide which arises from Ca(OH)2 <=> Ca 2+ + 2OH-, equilibrium law while noting that concentration of solid Ca(OH)2 isn't relevant). Until you create a saturated solution of calcium hydroxide, all the calcium hydroxide will form OH-.

Once again, exceptions apply. If you have aluminium hydroxide in a concentrated base solution, you will form Al(OH)4 -. Al(OH)3 is actually insoluble. Several metal oxides like those of zinc and iron also have this property. But don't worry about this for VCE.

[Mao]

EDIT: beaten...

How do diprotic substances work in relation to pH?

For instance, if I have excess H2SO4, I understand it won't fully ionise to SO4^2- + 2H+

that is, all of the first H+ will be removed, but not all of the second - is that right?

How do we calculate the pH then?

Also, do bases have the same effect? For instance if I have excess Ca(OH)2, will it fully ionise to Ca2+ + 20H-

or will the second hydroxide ion not be released from all of the calcium hydroxide?

Re: sulfuric acid, since it is such a strong acid, you can assume the first ionisation is complete. You can then use this information to solve for the equilibrium of the second ionisation (if Ka is provided). Note that this is different to your normal acid/base equilibrium:

.....(1)
.....(2)



Let the initial concentration of . Reaction (1) is complete, so gives us and . The equilibrium expression therefore becomes



Which is a quadratic equation.

As you can see, this is a lot harder than your usual acid/base equilibrium questions. It is extremely unlikely that VCAA would ask something like this.

For weak, polyprotic acids (e.g carbonic acid, phosphoric acid), the situation is even more complex. Solving these systems require non-linear systems of two or three variables, which are usually computed numerically rather than exactly (i.e. we can't easily write down a formula for the answer).


As for Ca(OH)2, it is unlike a diprotic base that can exist in several deprotonated states (i.e. 1-, 2-, 3-, etc). Ca(OH)2 is a salt, so for the sake of ionisation in pure water, the salt crystal is either dissolved (all OH- is free) or it remains as an undissolved crystal (OH- not free at all).

If, however, we were to add acid, the acid would react with the dissolved hydroxide, and also eat away at any undissolved crystal. This is why acid rain can erode limestone.

[Patches]

Ah, that makes sense - thank you both. Incidentally, where did you find all this out Nliu? I assume you're in year 12 - are you just going ahead and doing uni stuff already?