Solubility Question from a GS CBT

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Renaissance Man

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[FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif]Here is the question:

Given that the Ksp of FeX2 is 5.0 x 10-16 where "X" is an unknown anion, what is its solubility in moles per liter?

(the 2 in the salt above is supposed to be subscript)

I know that you have to find "s" by using 4s^3=Ksp, and if you do that, you will get s equals 5.0 x 10
.
.[FONT=Verdana, Arial, Helvetica, sans-serif]^-6 (and this is the correct answer according the Gold Standard)

However, isn't the solubility of "X" double this number?


.
 
The 4s^3 looks correct for finding the solubility of FeX_2. The concentration of X when FeX_2 dissociates into ions is 2s. I don't know if solubility's the right term for it though.
 
Solubility of FeX2 is "s", and the concentration of X is "2s". There is no such thing as solubility of X, per se, since solubility is a property of the solid.
 
Let's say you have a solid: PbI2. This solid dissolves into water like this: PbI2 ---> Pb + 2I.

Writing the Ksp expression will look like this: Ksp = [Pb]^2
The symbols in brackets represents the concentration of each ion.

To find molar solubility, you have to consider how the solid dissolves. Look again at the equation in bold.
For every 1 mole of "PbI2" that you have, you produce 1 mole of Lead and 2 moles of Iodide. We can represent the concentration of Pb as "X" and the concentration of 2I as "2X."

Substitute these values back into the brackets.

Ksp = [X][2X]^2
Ksp = 4x^3

Solving for "X" IS the molar solubility. Hopefully now you can understand the distinction between molar solubility and concentration.

Now if you wanted to find the concentration of I, you multiple the molar solubility twice ("X" x 2). That would be your concentration of Iodide.
 
Sorry, didn't see others responded. I had this window opened a while back and just now got to responding to it 😛
 
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