Aamc 10 ps #25

[MediCynical]

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Which of the following solutions has the lowest freezing point?

A) .10 M CaCl2
B) .15 M NaCl
C) .15 M CaCl2
D) .20 M NaCl

Answer: C


So my gut instinct told me C, but I changed it to D after calculating the ion product of each:

.15 M CaCl2
I.P. = [Ca2+][Cl-]^2 = (x)(2x)^2 = 4x^3 = 4(.15^3) = .0135

.2 M NaCl
I.P. = [Na+][Cl-] = (.2)^2 = .04


Does the ion product not have any bearing on freezing point depression/boiling point elevation?

If there was a problem in which 2 aqueous solutions were compared and the Ksp of each was given, would the solution with the higher Ksp have a lower freezing point/higher melting point?

I understand that there are 3 aqueous ions for CaCl2 and only two for NaCl, but I'm just confused with any bearing the IP/Ksp has on this.

 

[loveoforganic]

I really don't remember what ion product is, or if I've ever heard of it for that matter. At any rate, for colligative properties (one of which is freezing point depression), the only thing that matters is the number of particles in solution. CaCl2 at .15M puts more particles into solution than NaCl at .2M.

 

[mcatbeast]

you dont need to really use Ksp. just figure out the i (van hoft factor which says how many particles the compound will break into)
NaCl i= 2
CaCl2 i=3
NaCl i=2

then just multiply the molarity of each by the corresponding value for i.

A) .10 M CaCl2 3x.10= .30
B) .15 M NaCl 2x .15= .30
C) .15 M CaCl2 3x.15= .45 since c gave the greatest value thats the answer
D) .20 M NaCl 2x.20= .40