Ksp vavlue

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Hi Guys, I'm confused how they solve for Ksp valvue
Ksp = S(2s)^2
= S(4s^2)
Ksp = 4S^3

I don't get the second step how did they end up with S(4S^2)? and what happen to [M]?

Please help

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I had the same issue when I encountered a similar problem. Here is how it goes:

Suppose M is Barium in M(OH)2
BaOH2 --> Ba2+ + 2OH-

In that case, it's Ba(OH)2 and Ksp would be set up as follows:

Ksp = [Ba2+][OH-]^2
Ksp = [x][2x]^2
Ksp = x*4x^2
Ksp = 4x^3
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For every 1 M(OH)2 that are dissolved, 2 OH are present. So it is 2S, not just S.
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So I have a good question. Kaplan 2015 Chen book has a question regarding solubility:

"A saturated solution of cobalt(3) hydroxide (ksp = 1.6 * 10^-44) is added to a saturated solution of thallium(3) hydroxide (ksp = 6.3 * 10^-46). What is likely to occur?

A. Both cobalt hydroxide and thallium hydroxide remain stable in solution.
B. Some cobalt hydroxide precipitates while thallium hydroxide remains stable in solution.
C. Some thallium hydroxide precipitates while cobalt hydroxide remains stable in solution.
D. Both cobalt hydroxide and thallium hydroxide precipitate in solution."

The correct answer according to Kaplan is C. however I don't see how this is the answer because when you mix the solutions you are diluting both metal cations which will cause a shift to the left not the right for both the solutions. Yes common ion effect also pushes the solution to the left but how can we tell which "wins".

The TI(OH)3 solution is saturated, so adding more -OH will push the reaction to the left (precipitate). Co(OH)3 will remain stable as its Ksp is very low.

I think the TI and Co metal ions are distinct and can't be grouped together. Is this what you're referring to?
Well when you add the solutions you are diluting the metal ions. Regardless with metal ion concentration decreasing in a saturated solution a shift would occur to the right. I think the assumption of no solid material must be the only restriction from this occurring.