Chemistry 3/4 2013 Thread

[Jeggz]

Why do Concentrated Chlorine ions always oxidise preferentially to water? :\

Anyone?

[thushan]

Why do Concentrated Chlorine ions always oxidise preferentially to water? :\

The E value is dependent on concentration - so when you have concentrated [Cl-], instead of 1M [Cl-], the E value of Cl2 + 2e <--> 2 Cl- changes such that Cl- is preferentially oxidised to water.

The Eo values you see are given substances are at 1 atm pressure for gases and 1 M concentration for aqueous substances.

[Jeggz]

The E value is dependent on concentration - so when you have concentrated [Cl-], instead of 1M [Cl-], the E value of Cl2 + 2e <--> 2 Cl- changes such that Cl- is preferentially oxidised to water.

The Eo values you see are given substances are at 1 atm pressure for gases and 1 M concentration for aqueous substances.

Legend Thush
ThankyouU!

[lzxnl]

Think of it this way. If you had saturated NaCl dissolved in water and then you added KMnO4, there are two main competing equilibria; that of permanganate reacting with Cl- and with H2O. Under standard conditions of 1 M Cl-, MnO4- then it will prefer to oxidise water. However, if you have too much Cl-, then the equilibrium shifts to reduce the amount of Cl- there is.

[thushan]

Think of it this way. If you had saturated NaCl dissolved in water and then you added KMnO4, there are two main competing equilibria; that of permanganate reacting with Cl- and with H2O. Under standard conditions of 1 M Cl-, MnO4- then it will prefer to oxidise water. However, if you have too much Cl-, then the equilibrium shifts to reduce the amount of Cl- there is.

Nliu95 - VCE Chem doesn't explain that Eo values are a representation of the K value of a reversible reaction :/ your explanation, whilst practically correct, can lead to all sorts of misunderstandings.

[Patches]

Can anyone help explan #18 of multiple choice on the 2011 exam one? I've attached the question and the answer.

As far as I can see, it's impossible to conclude either A or B from the data book, since while the infrared data goes C-Cl,C-C,C-O,C-H , by either electronegativity or atomic mass the order should be C-Cl, C-O, C-C, C-H.

Can you get the answer from the data book, or is it just something you need to have known already? Cheers

[Alwin]

Can anyone help explan #18 of multiple choice on the 2011 exam one? I've attached the question and the answer.

As far as I can see, it's impossible to conclude either A or B from the data book, since while the infrared data goes C-Cl,C-C,C-O,C-H , by either electronegativity or atomic mass the order should be C-Cl, C-O, C-C, C-H.

Can you get the answer from the data book, or is it just something you need to have known already? Cheers

It's something you need to know, but once you know it its pretty "simple"
Since IR is all about vibrational energy (stretching, bending etc) the general rule is heavier atoms vibrate slower because of their greater mass. Hence, they have smaller wavelengths.
Also, clearly as you pointed out it is hard to see the relation in the data book, and that's coz other factors (namely bond strength) are important but mass is the most influential one out of options A-D.

Check out the examiners report too, usually they give explanations for MC qs.

hope it helped

[lzxnl]

There are a variety of factors at play here. Firstly, larger atoms have two effects. They're more massive, so any vibration effects would be smaller as noted. Secondly, larger atoms have larger bond distances, which lead to weaker attractions. Thirdly, larger atoms in the same period are actually smaller, so the bond distances may well be smaller. Fourthly, larger atoms in the same period are generally more electronegative, so if bonded to something not electronegative at all, like carbon, an electric dipole is formed and there is an ionic attraction as well, strengthening the bond.

I don't particularly like a black-and-white "it's larger, so it vibrates less" response. As an example, single C-O bonds have a maximum vibration wavenumber of around 1300 cm^-1, while H-O bonds can go up to 3670 cm^-1 according to http://en.wikipedia.org/wiki/Infrared_spectroscopy_correlation_table

Last time I checked, carbon was larger than hydrogen. So. Problem?

[thushan]

Let's simplify this - the wavenumber is dependent on two things: 1) mass; 2) bond length (which is dependent on many things, including the number of covalent bonds).

You should know that increased mass of atoms decreases the wavenumber at which a vibration stage is changed.
Shorter bond length (eg. a double bond instead of a single bond) will itself tend to increase the wavenumber.

Physical size of the atom would kind of matter as it would partly determine the bond length.

[Scooby]

VCAA 2012 Q3
c) - Wouldn't the HCl also react with H₂O, reducing its concentration, instead of just OH^-?

d) - Why is the percentage dissociation of the acid to two significant figures instead of three (ie. 50% instead of 49.9%)?

When we convert pH to H₃O⁺ concentration should we write that as
[H₃O⁺] = 1.00x10^-pH or [H₃O⁺] = 10^-pH?

For this question we couldn't assume that the initial concentration of the acid is equal to the equilibrium concentration.... so for what K_a values can we assume that [Acid]_initially = [Acid]_equilibrium and for what values can't we?

[Limista]

VCAA 2012 Q3


When we convert pH to H₃O⁺ concentration should we write that as
[H₃O⁺] = 1.00x10^-pH or [H₃O⁺] = 10^-pH?


You should write it as [H₃O⁺] = 10^-pH


so for what K_a values can we assume that [Acid]_initially = [Acid]_equilibrium and for what values can't we?

For Ka values < 10^-4, we assume it is a weak acid, so there are significant amounts of reactants in the equilibrium mixture, but not significant amounts of products. Therefore, concentration of weak acid at equilibrium is same as it was initially. Example is ethanoic or methanoic acid.

[lzxnl]

VCAA 2012 Q3
c) - Wouldn't the HCl also react with H₂O, reducing its concentration, instead of just OH^-?

d) - Why is the percentage dissociation of the acid to two significant figures instead of three (ie. 50% instead of 49.9%)?

When we convert pH to H₃O⁺ concentration should we write that as
[H₃O⁺] = 1.00x10^-pH or [H₃O⁺] = 10^-pH?

For this question we couldn't assume that the initial concentration of the acid is equal to the equilibrium concentration.... so for what K_a values can we assume that [Acid]_initially = [Acid]_equilibrium and for what values can't we?

For the first point, water is the solvent. As the number of moles of water drop, so does the volume of water. Also, the concentration of water in water is so ridiculously high in comparison that you can neglect water in aqueous equilibria generally.

There are two factors at play here. Firstly, the Ka value needs to be sufficiently high. Secondly, the concentration can't be too low, otherwise the percentage ionisation will be too high.

I'll give an example. Let's assume that we have a solution of concentration M molar, acidity constant K.
Assume that the concentration of H+ and conjugate base formed is x.
So M-x molar acid left.
Subbing into K expression
K = x^2/(M-x)
x^2 = MK - xK
x^2 + xK - MK = 0
x = 1/2*(-K + sqrt(K^2+4MK))                                    quadratic formula, plus sign because x > 0

We want the case when K << M as that is what normally happens

x = 1/2*(-K + 2*sqrt(MK)*sqrt(1 + K/4M))  by bringing out the sqrt(4MK) from the square root

People are going to hate me for doing this, but I'm going to go ahead anyway.
(1+a)^n, for small a, is roughly equal to 1 + an + n(n-1)/2 * a^2

So apply this to the square root term
sqrt(1+K/4M) = (1+K/4M)^1/2, and assuming K/4M is small, which it generally is:
sqrt(1+K/4M) ~ 1+K/8M + 1/2*-1/2*(K/4M)^2 = 1+K/8M - K^2/64M^2

Subbing this back into the original expression for x
x~1/2*(-K+2sqrt(MK)*(1+K/8M - K^2/64M^2))

OK. Let's see what this means. If M is sufficiently larger than K, then K^2/64M^2 is meaninglessly small. We can discard that term.
If K/8M is also sufficiently small, then we can remove that term as well.
We are then left with x~1/2*(-K + 2sqrt(MK) = -K/2 + sqrt(MK)
The term sqrt(MK) is what you would normally get by assuming M>>x in K = x^2/(M-x)
But if you assumed that K/8M was tiny, you wouldn't care about the individual value of K, so the -K/2 term is also negligible in this case.

What does all this mean? You need to look at the ratio K/M and consider how significant it is to the concentration in M of acid produced. Generally, if you have a concentration of around 0.1 M, and you have a normal weak acid with a Ka of 10^-5, then K/M = roughly 10^-4, which is tiny. But if you had a solution of 10^-3 M and your Ka was 10^-4...then K/M is 0.1, which can't be neglected.

[Scooby]

What does all this mean? You need to look at the ratio K/M and consider how significant it is to the concentration in M of acid produced. Generally, if you have a concentration of around 0.1 M, and you have a normal weak acid with a Ka of 10^-5, then K/M = roughly 10^-4, which is tiny. But if you had a solution of 10^-3 M and your Ka was 10^-4...then K/M is 0.1, which can't be neglected.

Is the acid we're talking about here hydronium?

And wouldn't reaction of HCl with sorbate ions also affect the shift in the equilibrium position?

[lzxnl]

Yes, concentration of H+ produced, or the value x that I was dealing with.


If you mean adding HCl to sorbate...reacting sorbate with acid to form sorbic acid and removing hydroxide to push the equilibrium to the right mean the same thing.

[Scooby]

So say we add HCl to the solution. The concentration of sorbic acid increases and the concentration of sorbate ions decreases. The concentration of hydroxide decreases. Would the equilibrium position shift to the left (to increase the concentration of sorbate ions) or to the right (to increase the concentration of hydroxide). That's where I'm confused. I mean I know it shifts to the right, but I'm not completely sure why