Why does adding solute particles decrease the freezing point?

[Borek]

As they are placed in the same container/system, I think that the vapor pressure will be equal over the brine as well as the ice.

Yes, but is this an equilibrium vapor pressure?

 

[Mr Real]

Yes, but is this an equilibrium vapor pressure?

If you mean if it'd be the same as the vapor pressure if the ice or the brine had been placed alone in the container, then I think it wouldn't. But if you didn't mean that, then I can't think why it isn't an equilibrium vapor pressure.

 

[Borek]

If you mean if it'd be the same as the vapor pressure if the ice or the brine had been placed alone in the container, then I think it wouldn't.

So now you wrote the pressures over brine and solid are different, but before you wrote they are identical:

I think that the vapor pressure will be equal over the brine as well as the ice.

We have already established, that if you have two samples with different equilibrium vapor pressures in the same container, mass will be transported from one sample to another till the system reaches the equilibrium (whatever it means).

 

[Mr Real]

So now you wrote the pressures over brine and solid are different, but before you wrote they are identical:

No, I meant that if we had placed either ice(with VP. v1) or brine(with V.P. v2 ) alone in the container then these would be different from the equilibrium vapor pressure we'd get if we placed both of them together.

 

[Borek]

OK, so now try to predict what will happen if you put a piece of ice and a beaker of brine, both at 0°C, in the same container.

I already told you to consider vapor pressures of both, please don't force me to move your legs one by one at each step.

 

[Mr Real]

OK, so now try to predict what will happen if you put a piece of ice and a beaker of brine, both at 0°C, in the same container.

I already told you to consider vapor pressures of both, please don't force me to move your legs one by one at each step.

Then, we would obtain the same vapor pressure over both of them.

 

[Borek]

Then, we would obtain the same vapor pressure over both of them.

Are the initial pressures identical?

 

[Mr Real]

Are the initial pressures identical?

No, the initial pressure is less over the ice compared to the brine (because ice is a solid whereas brine is a solid-liquid solution).

 

[Borek]

No, the initial pressure is less over the ice compared to the brine (because ice is a solid whereas brine is a solid-liquid solution).

Think it over.

At 0°C vapor pressure above ice and pure water are identical, which makes it possible for water and ice to coexist at the melting point.

What happens with vapor pressure over the water, when you add salt? Apply the Raoult's law.

 

[Mr Real]

What happens with vapor pressure over the water, when you add salt?

Oh, the vapor pressure of water will decrease on adding salt. So, the initial vapor pressure of ice will be greater than that of brine (seems very odd to me).

Apply the Raoult's law.

But I have read Raoult's law is applicable only for liquid-liquid solutions and this (brine) is a solid-liquid solution we are talking about, isn't that correct?

 

[Borek]

Oh, the vapor pressure of water will decrease on adding salt. So, the initial vapor pressure of ice will be greater than that of brine.

Yes. Now, remember that part about mass transfer when the equilibrium vapor pressures are different? Will the ice survive in such conditions, or will it evaporate and condense diluting the solution? In other words: is it possible for the ice to coexist with a brine?

But I have read Raoult's law is applicable only for liquid-liquid solutions and this (brine) is a solid-liquid solution we are talking about, isn't that correct?

Raoult's law works works for every mixture (well, for every ideal mixture, but it doesn't matter much here). However, as the vapor pressure over the solid is typically negligible, for liquid-solid mixtures we use it to calculate vapor pressure of the solvent only.

 

[Mr Real]

Yes. Now, remember that part about mass transfer when the equilibrium vapor pressures are different?Will the ice survive in such conditions, or will it evaporate and condense diluting the solution?

So, as the initial vapor pressure over the brine is lower than that over the ice therefore condensation will occur over the brine equalizing both vapor pressures and diluting the brine. So I know the ice will vaporize but I don't know whether it will survive because I don't know just what amount of the ice will evaporate(as the ice will disappear only when it evaporates completely and I don't know why that will happen).

In other words: is it possible for the ice to coexist with a brine?

According to me, it is possible.

the vapor pressure over the solid is typically negligible

Yeah, seems pretty weird to me that why did ice have more initial vapor pressure than the brine (I've got the reasoning but as I have said it seems very odd to me that a solid has greater v.p. than a liquid)

 

[Borek]

So, as the initial vapor pressure over the brine is lower than that over the ice therefore condensation will occur over the brine equalizing both vapor pressures and diluting the brine.

Right. That means ice disappears in the presence of brine, so as long as the brine is present there is no ice at the melting point, and to solidify brine we have to lower the temperature till the vapor pressure over the brine is equal to that over the ice, yes?

So I know the ice will vaporize but I don't know whether it will survive because I don't know just what amount of the ice will evaporate(as the ice will disappear only when it evaporates completely and I don't know why that will happen).

No matter how much ice evaporates, is - ever - the brine diluted infinitely, or is - always - salt concentration finite? And as long as the concentration is finite, is the vapor pressure over the brine equal, or lower to the vapor pressure over the ice (we are talking Raoult's law all the time here)?

Yeah, seems pretty weird to me that why did ice have more initial vapor pressure than the brine (I've got the reasoning but as I have said it seems very odd to me that a solid has greater v.p. than a liquid)

We are talking about two different solids here. Ice is a solid made of small, covalent molecules, which makes it quite volatile. Ionic salts (like NaCl) which are what is dissolved in a brine, have the vapor pressure many orders of magnitude lower. I guess you were told about latter being non-volatile at all, which is not entirely true. As I wrote earlier every reasonably stable substance has a gas/solid equilibrium line on the phase diagram. Sure, the vapor pressure can be negligible for most practical purposes, but it still doesn't make it zero.

 

[Mr Real]

Right. That means ice disappears in the presence of brine, so as long as the brine is present there is no ice at the melting point, and to solidify brine we have to lower the temperature till the vapor pressure over the brine is equal to that over the ice, yes?
No matter how much ice evaporates, is - ever - the brine diluted infinitely, or is - always - salt concentration finite? And as long as the concentration is finite, is the vapor pressure over the brine equal, or lower to the vapor pressure over the ice (we are talking Raoult's law all the time here)?

Sorry, but I didn't understand most of what you said. The things I have understood until now is that if we place ice and brine together, the initial vapor pressure of the brine will be lower than the ice and thus condensation will happen over it( so it will be diluted to some extent and the salt concentration will decrease), and some amount of ice will evaporate, and the ice will stop evaporating as soon as both vapor pressures become equal.
So, I don't get why ice will disappear completely or that as long as brine as present, why can't ice be present too?

 

[Borek]

the ice will stop evaporating as soon as both vapor pressures become equal.

At melting point, what should be the concentration of brine for the vapor pressure over it to be identical to the vapor pressure over ice?

 

[Mr Real]

At melting point, what should be the concentration of brine for the vapor pressure over it to be identical to the vapor pressure over ice?

I don't know but I'll make a guess: zero concentration? (I don't know the reason).

 

[Borek]

No need to guessing, just use the Raoult's law - use it to calculate solvent (water) pressure over the solution, using x for a molar fraction of the salt. If x <> 0 can the pressure over the solution be identical with the pressure over the ice at melting point?

 

[Mr Real]

No need to guessing, just use the Raoult's law - use it to calculate solvent (water) pressure over the solution, using x for a molar fraction of the salt. If x <> 0 can the pressure over the solution be identical with the pressure over the ice at melting point?

I think it can, because condensation will occur over the brine raising its vapor pressure thus equalizing it with that of the ice (some amount of ice will evaporate). I don't think x>0 will stop the two vapor pressures from getting equal, am I wrong?

 

[Borek]

Stop guessing, we are a point where things can be shown precisely - and quite easily. Start with the Raoult's law:

P = P_0(1-x)

Calculate what is x if P equals P_0.

 

[Mr Real]

Stop guessing, we are a point where things can be shown precisely - and quite easily. Start with the Raoult's law:

P = P_0(1-x)

Calculate what is x if P equals P_0.

For that x should be equal to zero, that is the brine must be infinitely diluted by condensation.

 

[Borek]

that is the brine must be infinitely diluted by condensation

Which - technically - is impossible, isn't it?

So, can the brine coexist with the ice at the water melting point?

 

[Mr Real]

Which - technically - is impossible, isn't it?

So, can the brine coexist with the ice at the water melting point?

Okay, by the equation it seems obvious that brine and ice can't coexist at the same temperature. But hadn't we earlier discussed, that condensation will occur and so the vapor pressures would become equal?

 

[Borek]

But hadn't we earlier discussed, that condensation will occur and so the vapor pressures would become equal?

Sure, trick is, water will move from ice to brine till there is no ice, and there will be only one vapor pressure - that of the brine. There will be no two different sources of the vapor.

Previous case was with two brine samples, so it was a bit different - I used that example just to make it easier for you to come to the conclusion mass can be transferred through the evaporation/condensation. Then once the concentrations became identical, process stopped. Now, the setup is different, so the final equilibrium is different as well.

 

[Mr Real]

Sure, trick is, water will move from ice to brine till there is no ice, and there will be only one vapor pressure - that of the brine. There will be no two different sources of the vapor.

Okay, so ice will disappear but I've seen that you've been emphasising the fact that it is happening at the melting point, so will ice completely disappear at melting point only, or at all temperatures? (I haven't seen the melting point play any role here so I think it will happen at every temperature, which is counterintuitive to me).

 

[Borek]

Melting point is particularly easy to understand, as it is the only point at which water and ice naturally coexist. At other temperatures predicting the outcome is a bit harder, as relative amounts of brine and ice start to play a role.

But in general we have come to the conclusion - just watching the vapor pressure is enough to explain, why brine has a lower melting point than pure water. That was the original question, wasn't it?

 

[Mr Real]

Melting point is particularly easy to understand, as it is the only point at which water and ice naturally coexist.

But didn't we just say that ice will completely disappear because of vaporization?

But in general we have come to the conclusion - just watching the vapor pressure is enough to explain, why brine has a lower melting point than pure water. That was the original question, wasn't it?

Actually, my original question was: why does freezing point decrease on adding solute to a solvent and why doesn't this decrease depend on the nature or the size of the solute particles?
Another doubt: will ice completely disappear at other temperatures also if kept in a container with brine? (Applying Raoult's law seems to suggest it will, as the formula doesn't have temperature in it)

 

[Borek]

But didn't we just say that ice will completely disappear because of vaporization?

In the presence of brine - yes. In the presence of water - no.

Actually, my original question was: why does freezing point decrease on adding solute to a solvent and why doesn't this decrease depend on the nature or the size of the solute particles?

For the same reason Raoult's law works the way it is. What matters are the molar fractions of the solute/solvent.

Another doubt: will ice completely disappear at other temperatures also if kept in a container with brine? (Applying Raoult's law seems to suggest it will, as the formula doesn't have temperature in it)

Depends on the temperature, initial brine concentration and relative amounts of ice and brine, no simple answer here.

 

[Mr Real]

Depends on the temperature, initial brine concentration and relative amounts of ice and brine, no simple answer here.

But the formula by which we got that the solution should be infinitely diluted (partial pressure formula in post #49) and as that cannot occur the ice will completely disappear; doesn't have any term for temperature, so how can it play a role here?
Now, if you've cleared the above doubt, so how would freezing point decrease, you've explained till that if we keep ice and brine in a container, the ice would disappear, so what does that have to do with the decrease in freezing point?

Thanks
Mr R

 

[Borek]

But the formula by which we got that the solution should be infinitely diluted (partial pressure formula in post #49) and as that cannot occur the ice will completely disappear; doesn't have any term for temperature, so how can it play a role here?

P₀ is a function of temperature, pressure of the saturated vapor over ice changes with the temperature as well (although at a different speed).

Now, if you've cleared the above doubt, so how would freezing point decrease, you've explained till that if we keep ice and brine in a container, the ice would disappear, so what does that have to do with the decrease in freezing point?

If there is no ice, only brine, have you reached the melting point?

What happens to the pressure of the saturated vapor over the ice when you lower the temperature? Does it go up, or down? If it goes down - what will happen when it reaches the value given by the Raoult's law for the given brine concentration (and P₀(T))?

 

[Mr Real]

My reasonings:

If there is no ice, only brine, have you reached the melting point?

If melting point is the point at which both solid and liquid states of the substance coexist, then if there is no ice that means melting point has not been reached.

What happens to the pressure of the saturated vapor over the ice when you lower the temperature? Does it go up, or down?

When the temperature is lowered, kinetic energy of the molecules decreases so vapor pressure of the ice decreases.

If it goes down - what will happen when it reaches the value given by the Raoult's law for the given brine concentration (and P0(T))?

If it reaches the value of P (which is the brine's vapor pressure), then it would coexist with the brine. But I've some doubts regarding this.
If the temperature is decreased, won't the pressure of the brine decrease too?
Also won't the Raoult's law prohibit these vapor pressures from becoming equal (like in the previous case) ?