Freezing Point and Ion dissociation

[RickP]

which of the following has the lowest freezing point.

0.15 M CaCl2
or
0.20 M NaCl

I was wondering if someone could explain the difference between the concentration vs the amount of dissociated ion species and how they can be viewed to know when the concentration is more influential then the number of dissociated ions!

Thanks

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[FeinMS]

which of the following has the lowest freezing point.

0.15 M CaCl2
or
0.20 M NaCl

I was wondering if someone could explain the difference between the concentration vs the amount of dissociated ion species and how they can be viewed to know when the concentration is more influential then the number of dissociated ions!

Thanks

Assuming that you meant 0.15 molality of CaCl2 and 0.20 molality of NaCl instead of M (Molarity,) Freezing point depression can be found by K*i*m. K is a constant for the solvent and assuming that same solvent is used, we can ignore that term. i refers to number of ions dissociated by dissolving the compound. For example, CaCl2 dissociates into 1 Ca2+ and 2 Cl-, so i=3 for CaCl2. Same thing for NaCl, Na+ and Cl-, so i=2. m is molality.

For CaCl2, i*m=3(0.15)=0.45. For NaCl, i*m=2(0.2)=0.4. Therefore freezing point will be more depressed by CaCl2. Note that actual value of depression is K*(0.45) and K*(0.4) respectively. I simply ignored K for the simplicity.

 

[RickP]

Assuming that you meant 0.15 molality of CaCl2 and 0.20 molality of NaCl instead of M (Molarity,) Freezing point depression can be found by K*i*m. K is a constant for the solvent and assuming that same solvent is used, we can ignore that term. i refers to number of ions dissociated by dissolving the compound. For example, CaCl2 dissociates into 1 Ca2+ and 2 Cl-, so i=3 for CaCl2. Same thing for NaCl, Na+ and Cl-, so i=2. m is molality.

For CaCl2, i*m=3(0.15)=0.45. For NaCl, i*m=2(0.2)=0.4. Therefore freezing point will be more depressed by CaCl2. Note that actual value of depression is K*(0.45) and K*(0.4) respectively. I simply ignored K for the simplicity.

ok I understand your explanation but it was M (molarity). Is molarity and molality interchangeable in terms of the relevant amount of moles within the solution?

 

[cheetoh]

You use molality when you're actually calculating a depression in freezing point - that isn't the goal here.

Tf = K * i * m (use molality here to calculate a change)

You aren't looking for a change; you're looking for the amount of ions in solution to see which would have more of a colligative effect.

 

[RickP]

You use molality when you're actually calculating a depression in freezing point - that isn't the goal here.

Tf = K * i * m (use molality here to calculate a change)

You aren't looking for a change; you're looking for the amount of ions in solution to see which would have more of a colligative effect.

Thus coming full circle to my original question to discern between the effect of amount of ions vs concentration. Do they have an effect on each other or are they independent.

 

[FeinMS]

Thus coming full circle to my original question to discern between the effect of amount of ions vs concentration. Do they have an effect on each other or are they independent.

They're independent of each other but both contribute to freezing point depression. Note i=amount of ions and molality=concentration.

 

[DrBTS]

Thus coming full circle to my original question to discern between the effect of amount of ions vs concentration. Do they have an effect on each other or are they independent.

I understand your question, and the answer is they are independent properties but both fall under the umbrella of properties that are effectors of 'amount' of substance. So you can probably figure out that an increase in molarity of one substance may have a more significant effect on FPD than ion dissociation constant, if it was concentrated enough. In this case, if we have the twice the molar weight w/ CaCl2, and thus greater molality, not to mention 3 dissociated ions vs 2 so you can make a fairly confident estimate that the former has greater FPD value. But quite honestly, assuming you understand my logic, you should probably do the math and please let me know if this is in fact not the case.

 

[sakabato93]

They're independent of each other but both contribute to freezing point depression. Note i=amount of ions and molality=concentration.

Remember that i (van't Hoff factor; gotta know the name) is the amount of dissociated ions per mole of ionic compound dissolved. Its not simply the amount of ions. Insoluble ionic species like AgCl still contain ions, but the van't Hoff factor is practically zero (actually is around 10^-6 since a tiiiiny bit will dissolve).