Unit 4 Questions MEGATHREAD!

[Mao]

D is wrong.

Liquids and solids are NEVER included in an equilibrium expression. The reasons are to do with something called 'activity.'

You're right HenryP it's B.

This is only partially correct. In most applications, you would be correct. However, approximating the activity of pure liquids and solids is always an approximation.

But otherwise, B is the most pragmatic answer.

[thushan]

Yup, point taken

[paulsterio]

Since molarity is the number of moles per litre, does that mean that the concentration of water can be found by

(1kg H2O / 1 L) * (1 mol H2O/ 18 gm) = (1000/18) mol/L = 55.56 M

Can't we technically find the "concentration" of anything knowing its density?

[b^3]

Since molarity is the number of moles per litre, does that mean that the concentration of water can be found by

(1kg H2O / 1 L) * (1 mol H2O/ 18 gm) = (1000/18) mol/L = 55.56 M

Can't we technically find the "concentration" of anything knowing its density?

Yeh, think and , all you are doing is subbing that in. And the result you get is basically the same (I think it is 56M) that is given for water in the Heinemann book, when they are explaining Ka values.

so

EDIT: added rearrange formula.

[paulsterio]

Thanks b^3

But out of curiosity, say you have 1M NaCl, 1L of it is in a beaker and you add 1L of Ethanol, does that change the concentration of NaCl?

[b^3]

The voulme is changing while the amount of NaCl is the same so it should change the concentration of NaCl.
Upon further reading NaCl dissolves very little in ethanol so I think we are going to need Thushan or Mao to answer this.

[b^3]

I was just thinking that.

[paulsterio]

Upon further reading NaCl dissolves very little in ethanol so I think we are going to need Thushan or Mao to answer this.

Yeah, that's why I was wondering, because Ethanol and Water are miscible, but then NaCl is only really soluble in the Water :S

[b^3]

Also thought I'd point out an interesting situation.
If you add 50 mL of distilled water to 50 mL of ethanol you end up with a solution with a volume of 98 mL because of the intermolecular forces.

So the conc of water here would be



Does that makes sense?

[taqi]

For the purposes of the exam, do we assume strong acids are 100% ionized? So if I were to add water too it, ionization would not increase because its already complete. Or do we assume its still got a very high K value, and is 99% ionized, so adding water does increase ionization?

[thushan]

Well, given that strong acids have Ka values of about 10^6, and a 1M solution of an acid with a Ka value of 10^6 is 99.9999% (that actual value) ionised, you can assume that no further ionisation occurs and that reaction is complete upon addition of water.

[b^3]

But out of curiosity, say you have 1M NaCl, 1L of it is in a beaker and you add 1L of Ethanol, does that change the concentration of NaCl?

While your here Thushan and before this gets lost, do you have any ideas on this?

[thushan]

Thanks b^3

But out of curiosity, say you have 1M NaCl, 1L of it is in a beaker and you add 1L of Ethanol, does that change the concentration of NaCl?

OK. I think i'll explain the whole thing. Mao, correct me if I make any errors.

When we work out K values, we look at something called the "activity" of the species. For the sake of formatting, let a(A) be the activity of A.

In reality, say we had a reaction aA + bB <--> cC + dD

K = a(D)^d a(C)^c / a(A)^a a(B)^b. Familiar right?

Ok, so now how do we measure activity of a species?

For GASEOUS and AQUEOUS species a(A) is approximately equal (esp at lowish concentrations) to [A], so for an aqueous or gaseous species, we can replace activity with concentration (molarity).
For SOLID and LIQUID species, it so happens that their activities, regardless of their density or "concentration" or whatever, are around 1.

So say we have HA + H2O <--> H3O+ + A-

K = a(A-)a(H3O+)/a(HA)a(H2O) = [A-][H3O+]/[HA] x 1 = [A-][H3O+]/[HA]. Familiar?

That's why solids and liquids don't appear in the equilibrium expression.]

So, really, working out concentrations of liquids and solids, whilst theoretically possible, is irrelevant to equilibria, given that their activity remains at 1.

@b^3: Woah, no idea what happens there. Will pass on to Mao! I would think that if NaCl is soluble in the mixture, then yes its concentration would change. But this is beyond my knowledge. MAO!

[Mao]

This is very tricky. You guys are talking about three different topics at once... But let me try to clear up some things.

But out of curiosity, say you have 1M NaCl, 1L of it is in a beaker and you add 1L of Ethanol, does that change the concentration of NaCl?

While your here Thushan and before this gets lost, do you have any ideas on this?

Firstly, salt is sparingly soluble in ethanol (~0.01 M). Literature suggest in a 50:50 mixture of water/ethanol, the solubility of NaCl may be as high as ~0.1M

The definition of 'concentration' is relative. In a system, you would define something as the solvent, and something as the solute. The concentration would thus be the amount of solute per amount of solvent. If you wish to define the 50:50 mixture of water/ethanol as the solvent, then you can certainly calculate NaCl concentration from that. (In the hypothetical case, it will be 0.5M.)

We generally define water as the solvent, so concentrations of aqueous solutions are quite standard. Due to various changes to density upon mixing solvents, this stops being so clear cut.

When we talk about dissolution, we generally mean small amounts of solute dissolving in large amounts of solvent.
When you talk about mixing 1L of water with 1L of ethanol, you are more talking about the miscibility of the two solvents. The situation gets much more complex.


What about activity?

When I mentioned to Thushan before that activity of pure liquids is only approximately 1, I had this in mind.

Activity of water in dilute aqueous solutions is approximately 1, because it is constantly colliding with the solute.

When we change the solvent to 50:50 water:ethanol, the activity of water will not be 1. It will be quite different, because the solute will no longer be constantly colliding with water. It is quite difficult to predict what this activity will be.

Also, . In fact, . Each species have a different constant of proportionality (called activity coefficient), but it is a constant nonetheless. This means the concentration quotient is simply a scaled version of the activity quotient (the real equilibrium constant), and so for convenience sake we just use the concentration quotient.

Note that this equation is only valid for working out the 'molar density' of pure solvents.


I am tired. I will write a more coherent explanation tomorrow.

[mickeymouse]

um could someone help explain vcaa 2010 mc q8?

why the hydroxide ions increase when diluting HCl ...