March 03, 2024, 04:23:00 AM
Forum Rules: Read This Before Posting

### Topic: Feezing point depression theoretical calculations  (Read 3346 times)

0 Members and 1 Guest are viewing this topic.

#### ale1998

• New Member
• Posts: 5
• Mole Snacks: +0/-0
##### Feezing point depression theoretical calculations
« on: December 11, 2016, 09:29:15 AM »
Hello,
I did a lab measuring the freezing point depression of the different salts (BaCl2, CaCl2, ZnCl2, MgCl2, SrCl2) at different concentrations in demi water. I need to calculate the theoretical values of the freezing point depression to my results. Since im using high concentrations of these salts i cannot treat the solutions as ideal solutions and use the following formula: ΔTF = KF · b · i
Where ΔTF is freezing point depression, KF is cryoscopic constant, b is the molarity, i is the van 't Hoff factor.

I will need to use the formula that takes into account the properties of the salt at high concentrations. This formula would be the one made by Ge and Wang on this wikipedia page:
https://en.wikipedia.org/wiki/Freezing-point_depression#Calculation

When i try to calculate the freezing point depression with this formula concentrations of above 1mol/dm^3 I get a negative value in the root. I tried to fix this problem by making ΔCfusp  have a value of -2.11 J/(g*K) instead of 2.11 J/(g*K) (I made the value negative). This eliminates the problem of the formula not working for higher concentrations and the values it gives seem more reasonable. However they are still quite far from my results (predicting a freezing point of 45K lower than my experimental results). This might be due to the osmotic coeficient that i need to use to calculate ln(aw) (where aw=activity of the solution). However these osmotic coeficcients are very difficult to find.

Does anybody maybe know where i can find the osmotic coeficients for the salts i am using? Or does anybody know why the value under the root of the formula turns negative?

If anybody wants to take a look at my results and calculations i will happily send you them!

Alessandro
« Last Edit: December 11, 2016, 09:58:39 AM by ale1998 »

#### mjc123

• Chemist
• Sr. Member
• Posts: 2046
• Mole Snacks: +296/-12
##### Re: Feezing point depression theoretical calculations
« Reply #1 on: December 12, 2016, 04:47:03 AM »
How can you get a negative value in the root? All the quantities are positive.

You had better post your calculations so we can see what you are doing.

#### ale1998

• New Member
• Posts: 5
• Mole Snacks: +0/-0
##### Re: Feezing point depression theoretical calculations
« Reply #2 on: December 12, 2016, 09:00:10 AM »
I get negative value under the root because ln(aliq) is negative when calculated with the following formula:
ln(aliq)=-(v*m)*φ/55.51
The link of this formula: https://en.wikipedia.org/wiki/Activity_coefficient#Theoretical_calculation_of_activity_coefficients
Where v is the number of ions produced from dissociation of one molecule of the dissolved salt, m is the molality of the salt dissolved in water, φ is the osmotic coefficient of water, and the constant 55.51 represents the molality of water.
since none of the constants (v, m or φ) will never be negative, the value of ln(aliq) will always be negative by definition.

This is a calculation for one of my results:
There is a concentration of 1.084mol dm-3 of BaCl2 in water.
Finding ln(aliq):
v=3
m=1.084
φ=1    (i don't know the value of the osmotic constant so i assume it is 1)
With these values:
ln(aliq)=-(3*1.084*1)/55.51=-0.0586

Knowing ln(aliq) we can apply the freezing point depression formula by Ge and Wang (this formula can be found here: https://en.wikipedia.org/wiki/Freezing-point_depression#Calculation)

The rest of the constants are:
ΔHfusTf=333.6 J/g
R=8.314 J/(K*mol)
Tf=273.15
ΔCfusp=2.11 J/(g*K)

Plugging these values in the root we get:
[2*ΔCfusp*Tf2*R*ln(aliq)+(ΔHfusTf)2](1/2)

=[2*2.11*273.152*8.314*(-0.0586)+333.62](1/2)

=[-42110.03](1/2)

This is a negative value and therefore the formula can't work for these values.
The only possible solution I think is that ΔCfusp is a negative value. Therefore i retried the calculation plugging the values in the full formula, with ΔCfusp=-2.11  J/(g*K) and it works. The value i get is -65.54K (this is how much the freezing point is depressed in that solution).
This value however is very far from the experimental value which is about -6K freezing point depression.

From here I think the only thing that could change the theoretical value is the osmotic coefficient (φ) which i do not know the values for.
However possibly i am doing something else wrong when calculating, or does anybody maybe know other possible reasons why the theoretical value is so different from the experimental value?

Hope this makes it more clear

Alessandro

#### mjc123

• Chemist
• Sr. Member
• Posts: 2046
• Mole Snacks: +296/-12
##### Re: Feezing point depression theoretical calculations
« Reply #3 on: December 12, 2016, 11:14:44 AM »
I'm not quite clear what's going on here - I'll review it in more detail when I have time - but a couple of things leap out:

m is molality, not molarity (they won't be hugely different)
You are mixing quantities in J/g with quantities in J/mol. Convert everything into "per mol" units and see what you get.

#### ale1998

• New Member
• Posts: 5
• Mole Snacks: +0/-0
##### Re: Feezing point depression theoretical calculations
« Reply #4 on: December 14, 2016, 12:55:46 PM »
In fact that was the problem and it is now fixed.
Now i have a function for freezing point depression of water at different concnetraions of salts that  dissociation into 3 ions from one molecule of the dissolved salt. This function is the same for the different salts (BaCl2, CaCl2, MgCl2, ZnCl2, SrCl2) because they all disociate in 3 ions and i did not take into account the osmotic coefficients because they were to difficult to find.
The function i obtain predicts a decreaseing freezing point of up to -96K at 14.0 mol dm-3. After any concentration of higher than 14.0 mol dm-3 results in imaginary numbers.

Is this maybe because the formula doesn't work for concentrations higher than 14mol dm-3? Because i read that the formula doesn't work well for high ionic concentrations.

#### mjc123

• Chemist
• Sr. Member
• Posts: 2046
• Mole Snacks: +296/-12
##### Re: Feezing point depression theoretical calculations
« Reply #5 on: December 14, 2016, 01:17:58 PM »

#### ale1998

• New Member
• Posts: 5
• Mole Snacks: +0/-0
##### Re: Feezing point depression theoretical calculations
« Reply #6 on: December 14, 2016, 03:24:36 PM »
I have read around about the Eutectic solutions and am starting to understand it.
But i do not see how this answers the point that the formula doesn't work anymore at concentrations of higher than 14 mol dm-3 (since imaginary numbers are formed)?

#### mjc123

• Chemist
• Sr. Member
• Posts: 2046
• Mole Snacks: +296/-12
##### Re: Feezing point depression theoretical calculations
« Reply #7 on: December 15, 2016, 05:10:33 AM »
What are the assumptions in your formula? Are they still valid at such high concentrations? Especially for ionic solutions? How much is 14 mol/L? That's more than half the volume. Can the solution be described as solvated ions in water? The ions will be scarcely separated. It's more like a solution of water in salt (which is what the phase diagram says it is). I can hardly think any theoretical equation will accurately describe it.