## Outline 1 Learning Objective [ENDORSED]

Samantha Pedersen 2K
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### Outline 1 Learning Objective

One of the learning objectives from Outline 1 says, "Use equilibrium constants to predict solubility." I'm guessing that if the equilibrium constant is large, then the compound is very soluble. If the equilibrium constant is small, then the compound isn't very soluble. Is this correct? Is there anything else we should know for this learning objective?

Shrey Pawar 2A
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### Re: Outline 1 Learning Objective

I would think that is all we need to know. I don't think we went over the solubility in any other context other than the mixing of the substances that can be shown by Kc. Hope this helps!

Irene Chang 2F
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### Re: Outline 1 Learning Objective

Yes, I believe you are correct. The larger the K, the higher the solubility since there are more products being formed.

Samuel Flores 1E
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### Re: Outline 1 Learning Objective

You are correct. If K has a large value, then this means that the equilibrium will sit to the right. This means that there will be a high concentration of products at equilibrium, which means the reaction is readily soluble, simply because the K has a large value.

Hope this helps!

Alara Aygen 3K
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### Re: Outline 1 Learning Objective

Yes! A high K value indicates that the reaction favors the products, which means it is more soluble.

Chem_Mod
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### Re: Outline 1 Learning Objective  [ENDORSED]

Correct. You can think about solubility as the reaction

$AB_{(aq)}\rightleftharpoons A^{+}_{(aq)}+B^{-}_{(aq)}$

and the expression for K:

$K=\frac{[A^{^{+}}][B^{-}]}{[AB]}$

Thus if K is large, there are many dissolved ions compared to the amount of undissolved species. If K is small, there is more undissolved material compared to the concentration of dissolved ions.