[Hunt]

We were required in class to solve the following problem:

The freezing pt depression of 0.091 m solution of CsCl is 0.302 degrees celcius. The freezing pt depression of 0.091 m CaCl₂ is 0.44  degrees celcius.

In which solution does ion-association appear to be greater? Explain.

I first computed the theoretical i ( the vant hoff factor )
For CsCl:
i(theo) = 2  , a biionic compound
i(exp) = ?T / K_bm = 1.78

Obviously the difference is due to ion-pairing. I computed ?i = 0.22

Thus, % ion association = 22%

For CaCl₂
i(theo) = 3  , a triionic compound
i(exp) = ?T / K_bm =2.59

I computed ?i = 0.41

Thus, % ion association = 41%

However, it turned out my answer is not entirely right. For CsCl , the answer was right, but for CaCl₂ , my computation was wrong.

This is how our Chemistry professor solved it:

He computed the i(theo) and i(exp) for both ionic compounds, same as above. Then:

For CsCl
We assume that i = 2 - 1 -----> 100% ionization
                         1.78 - 1 -----> ? = 78% ionization

Thus, % ion association = 22%

For CaCl₂
We assume that i = 3 - 1 -----> 100% ionization
                        2.59 - 1 -----> ? = 79.5% ionization

Thus, % ion association = 20.5%

My question is, why did he take into account the ion pairing by substracting 1 from i(theo) and i(exp) ? Why 1?

For Calcium chloride, there r 3 ions. Hence, there are C₃² = 3!/2!(3-2)! = 3 types of ion-pairs that could result not 1!                                

 

[Hunt]

I think this was my 2nd post on chemical forums. Now I can say I somehow understood why my professor was right although I do not like his method. Some time ago I found a general equation for computing the degree of association on wikipedia, which made me think about it and then I got it. For those interested, the idea is simple. Assume there's a certain ionic molecule that undergoes dissociation,

ABCD.... <---> A + B + C + D + .

This can be represented as A_v <---> vA

where v = number of species present.

for example, for CH₃COOH <----> CH₃COO^- + H⁺ , v = 2

If N particles of A_v are present then vN particles must exist for 100% dissociation. If association occurs, the number must be less than vN. Let "a" denote the fraction of association then the number of particles associated is = aN . Number of particles dissociated = vN(1-a)

Total number of particles in solution = number of particles associated + number of particles dissociated

N_T = aN +  vN(1-a) = N[v(1-a) + a]

i = N_T / N = v(1-a) + a

i= v + a(1-v)

This formula can be manipulated in many different ways. On wikipedia they substitute the association fraction 'a' for the dissociation fraction 'd'.

a + d = 1

i = v +(1-d)(1-v) = 1 - d + vd = 1 + d(v-1)

This can be re-arranged to get :

d = (i - 1)/(v-1)

http://en.wikipedia.org/wiki/Van_%27t_Hoff_factor

For the problem I posted earlier, it can now be solved with simple substitution. For CaCl2 , v = 3.

i = 2.59 then % a = 100 (i - v)/(1-v) = 2.59 - 3 / 1-3 = 20.5%

I still remember asking my profesor if the vant hoff factor could be related to the equilibrium constant. Now it seems pretty clear. Say for a monoprotic acid, Ka can be found in terms of the degree of ionization ad so in terms of 'i' by simple substitution.

[Rabn]

Couldn't you just use Ksp to find % ion association? I.e. for something with  a very small Ksp the ion association % would be very high. 

[Hunt]

Ofcourse you're correct. Ksp is used for precipitation reactions only , but yes in general K can be used to find the % of association or dissociation. However, the idea here is how to relate 'i' to the percent of dissociation/association. I did not get the right answer before with calcium chloride bcause I failed to understand the meaning of 'i' : "the number of particles of solute actually in solution per particle of solid solute added".

 

[Rabn]

Gotcha...I hope others read your thread as I'm sure that the matter of "i" may be one of those little things that can cause many big problems.