van't hoff factor

[Oh_Gee]

From GS-5 PS #5

"What would the freezing point in Kelvin of a solution which is 0.50 molal in sucrose and 0.50 molal in acetic acid be?
(Kf of water = 2.0 oC mol-1 and freezing point of water = 0 oC)

a. -1 K
b. -2 K
c. 272 K
d. 271 K

Answer D
From the equation Tf = Kfm, Tf = (2.0 oC mol-1)(0.5 molal + 0.5 molal) = 2.0 oC. Therefore, the freezing point of this solution is 0 oC - 2 oC = -2 oC = (273 - 2) K = 271 K.

{Recall: acetic acid is a weak acid thus relatively very little dissociates; CHM 6.1, 6.6}"


why is the van't hoff factor not taken into account here? sucrose (i=1) is not an electrolyte and i know acetic acid (i=1 probably?) dissociates little. so shouldn't delta T = 2 x 2 x (.5 +.5)?

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[techfan]

I think that since it is a mixture of the 2, and once one compound mixes it isn't pure water anymore so you'd have a different Kf. Here they are assuming since neither compound disassociates very much you can treat them as the same and assign a van't Hoff factor of 1 to the entire mixture.

 

[Oh_Gee]

I think that since it is a mixture of the 2, and once one compound mixes it isn't pure water anymore so you'd have a different Kf. Here they are assuming since neither compound disassociates very much you can treat them as the same and assign a van't Hoff factor of 1 to the entire mixture.

but then doesn't that mean if you were to add a strong acid like HCl to water, it's no longer water, so the kf will then be different and i=2? that means everytime if we were to calculate a freezing or boiling point elevation/depression, i would always be 1 or the kf/km would have to change?

 

[Czarcasm]

From GS-5 PS #5

"What would the freezing point in Kelvin of a solution which is 0.50 molal in sucrose and 0.50 molal in acetic acid be?
(Kf of water = 2.0 oC mol-1 and freezing point of water = 0 oC)

a. -1 K
b. -2 K
c. 272 K
d. 271 K

Answer D
From the equation Tf = Kfm, Tf = (2.0 oC mol-1)(0.5 molal + 0.5 molal) = 2.0 oC. Therefore, the freezing point of this solution is 0 oC - 2 oC = -2 oC = (273 - 2) K = 271 K.

{Recall: acetic acid is a weak acid thus relatively very little dissociates; CHM 6.1, 6.6}"


why is the van't hoff factor not taken into account here? sucrose (i=1) is not an electrolyte and i know acetic acid (i=1 probably?) dissociates little. so shouldn't delta T = 2 x 2 x (.5 +.5)?

You should probably treat them independently to avoid error. Collectively, we have 0.5m sucrose (i=1), for a total of 0.5m; this times Kf = 1 degree C. We also have 0.5m acetic acid (i=1, because it's also a molecule), for a total of 0.5m. This also times Kf = 1 degree C. Colligative properties tell us solutes decrease melting/freezing point, so collectively, a degree decrease in celcius = -2 degrees C. 273K + (-2) = 271K.

If you prefer your way, you could have also realized, collectively, there is 1m of stuff (not 2 as you used in your equation), times Kb (which is 2), and this would also equal 2. Either way, works. Just a simple math error.

Also, van't hoff factor is usually at minimum 1. Anything less would contradict what colligative properties tell us. NaCl for example, i=2. Even if you considered that acetic acid might dissociate a little bit, i would be a little more than 1, which would mean that the answer would be a little less than 271K (which is not even a choice).

 

[Oh_Gee]

You should probably treat them independently to avoid error. Collectively, we have 0.5m sucrose (i=1), for a total of 0.5m; this times Kf = 1 degree C. We also have 0.5m acetic acid (i=1, because it's also a molecule), for a total of 0.5m. This also times Kf = 1 degree C. Colligative properties tell us solutes decrease melting/freezing point, so collectively, a degree decrease in celcius = -2 degrees C. 273K + (-2) = 271K.

If you prefer your way, you could have also realized, collectively, there is 1m of stuff (not 2 as you used in your equation), times Kb (which is 2), and this would also equal 2. Either way, works. Just a simple math error.

but for my way, .5molal + .5 molal=1 molal which is what i put 2 x 2 x 1=4. where did i go wrong in my calculation?
wait do you add or multiply the van't hoff factors of acetic acid and sucrose

although i see the first way you did it is better than mine

 

[techfan]

Yeah, ignore my post, Czar has it correct, I kept thinking the freezing point of water was 272.3 instead of 273.2 so I tried to explain it away

 

[Czarcasm]

but for my way, .5molal + .5 molal=1 molal which is what i put 2 x 2 x 1=4. where did i go wrong in my calculation?
wait do you add or multiply the van't hoff factors of acetic acid and sucrose

although i see the first way you did it is better than mine

That's were conceptual understanding might help. The whole purpose of van't hoff factor is to help us calculate the amount of stuff of some solute in solution. Using your way, if you are summing the solutes together, you are treating them as if they are one unique substance and so, for this combined unique substance, since it does not dissociate, i=1.

If you can learn to disregard i all together and realize what it's useful for, it could avoid unnecessary error. Ultimately, we want to know the total molality of solutes. If you have 0.5 molal of NaCl and 0.5 molal of Sucrose, how much stuff do you have? Instead of thinking of terms of what i will equal, realize NaCl (a soluble salt), will dissociate completely. So, the total molality of NaCl is not 0.5, but 2x0.5 = 1m. Also, sucrose doesn't dissociate, so sucrose = 0.5m. Collectively, how much stuff do we have? 1.5m, and from there, you multiple the stuff you have by the constant to determine how that will effect temperature.

 

[Oh_Gee]

That's were conceptual understanding might help. The whole purpose of van't hoff factor is to help us calculate the amount of stuff of some solute in solution. Using your way, if you are summing the solutes together, you are treating them as if they are one unique substance and so, for this combined unique substance, since it does not dissociate, i=1.

If you can learn to disregard i all together and realize what it's useful for, it could avoid unnecessary error. Ultimately, we want to know the total molality of solutes. If you have 0.5 molal of NaCl and 0.5 molal of Sucrose, how much stuff do you have? Instead of thinking of terms of what i will equal, realize NaCl (a soluble salt), will dissociate completely. So, the total molality of NaCl is not 0.5, but 2x0.5 = 1m. Also, sucrose doesn't dissociate, so sucrose = 0.5m. Collectively, how much stuff do we have? 1.5m, and from there, you multiple the stuff you have by the constant to determine how that will effect temperature.

still confused. would i=3 in your example?
wait nvm i think i get it
so it's best to just calculate the delta T for each individual thing?

 

[Oh_Gee]

i have another question about this stuff from the same test. so i'm thinking i would still be 1 since i'm assuming sucrose degrades to glucose/fructose, both of which aren't charged. but how would the molality increase? is it because now there's 1mol fructose:1mol glucose?
i feel like i would need specific numbers to see if the freezing point changed

"The freezing point of a solution of sucrose was noted and then a dissacharidase, which degrades sucrose, was added to the solution. The enzyme was subsequently removed. How would the freezing point of this final solution compare to that of the original solution?


a It would be lower.
b It would be higher.
c It would be the same.
d It cannot be determined from the information given.

Sucrose would be degraded to smaller molecules by the enzyme. Thus the number of molecules in solution would increase, leading to an increase in the number of moles of molecules. The molality of the solution would then increase [Molality = (Number of moles of solute) / (1000 g solvent)]. From Tf = K'fm, the freezing point depression would increase and hence the freezing point would decrease. {This problem is a classical example of a colligative property which depends on the number not the type of molecule or particle present}"

 

[Teleologist]

Also, van't hoff factor is usually at minimum 1. Anything less would contradict what colligative properties tell us. NaCl for example, i=2. Even if you considered that acetic acid might dissociate a little bit, i would be a little more than 1, which would mean that the answer would be a little less than 271K (which is not even a choice).

Contradict? When we think of the van't hoff factor, we think about everything in idealized terms. NaCl dissolves completely in water. Sucrose doesn't dissolve at all in water. Etc. So NaCl has i = 2 because for every unit of NaCl we put in water we theoretically get two completely independent particles in solution: Na(+) and Cl(-).

The reality is much messier than that. Ion-pairing effects often screw around with us. The actual i factor of NaCl in water solvent is probably slightly <2 because of the ion-pair effect. Ion-pairing is probably touched on in orgo when dealing with SN1 mechanisms in solution; the leaving group doesn't necessarily fully dissociate from the substrate; ion-pair effects are responsible for the fact that many SN1 reactions don't necessarily yield a racemic product.