Ksp confusion..help

[dentalstudent2021]

ex: BiI3 (s) <-> Bi3+ + 3I-
Chad says Ksp =[Bi] [I-]^3 ...no 3 inside the brackets!

Then to find Ksp, he does,
Ksp = (x)(3x)^3 ???

confused me completely on Ksp. whats the difference here

how about on this Q:
2 A(g) + B(s) ⇌ 3 C(g) The above reaction is at equilibrium at STP and the molar concentrations found are [A] = 2.0 M, B = 2.0 mol, and [C] = 4.0 M. What is the equilibrium constant for this reaction at STP?

I said Ksp = [C]^3 / [A]^2 ...which is correct. Please help me with the differences here.

Thanks.

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[Lnguyen7]

For a general reaction
aA + eE <-> cC + dD
The equilibrium constant will always have the form: K = [C]^c[D]^d/([A]^a[E]^e) ((I want to use bB instead of eE, but somehow it screws up the format))
so for BiI3(s) <-> Bi3+ + 3I-
Ksp = [Bi3+][I-]^3

However, when you do the calculation, you put the concentration of I at equilibrium into [I-] , which is 3x.
BiI3(s) <-> Bi3+ + 3I-
I:..................0.........0
C:................+x.......+3x
E:.................x..........3x

Plug x and 3x as the concentration of Bi3+ and I- respectively into the expression, you then have Ksp = [Bi3+][I-]^3 = (x)(3x)^3.

 

[omarski]

I think you're missing the point. Fundamentally the idea of Ksp is to give you a value of how much of the solid you can dissolve in solution before a precipitate forms. This has to be accounted for the in Ksp equation that you make for each individual reaction/problem. The 3x and x are the ratios in comparison to the original solid. Because the original BiI3 has 3 Iodines for every molecule of BiI3, when the solid breaks down to it's ionic compounds, Iodine does not stay as 1 molecule of I3; rather, it becomes 3 molecules of Iodine. The ice chart demonstrates this very idea. The change would be a 1 to 3 ratio, hence why we add 3x and not x.

Secondly as chad highly recommends write out the ksp equation so these sort of confusions don't occur:
Ksp = [Bi3+][I-]^3
From there make an ice chart and solve for whatever the question asks for .
Also remember, what is x? Molar solubility. The reason the second equation is different is because the molar concentrations are given, so you don't have to solve for them. The first problem you gave does not have actual concentrations given, hence why x is used in the ice chart.

I highly recommend understanding the fundamental concept of whatever topic you're learning as it'll make your life much easier. There are many different types of questions for this particular topic and it can be incredibly difficult if you don't understand "why" everything is the way it is.