Ksp question about coefficients and exponents

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losinov

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Hey everyone,
Just going through equilibrium and acid/base stuff and I can't seem to understand how I know when to add the coefficient or just use the exponent when writing out the equilibrium equations.

For example:
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2

F remains with just an exponent of the 2 it doesn't become a coefficient.

In another problem:
MX2(s) → M2+(aq) + 2X-(aq)
Ksp = [M2+][X-]2 = (x)(2x)2 = 4x3
The X has 2 molecules similar to the BaF2 above but in this problem the 2 is an exponent and a coefficient.

I'm sure this is just a base rule that I am forgetting or something basic. Thanks for your help in advance everyone!

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Here's another example where BaF2 is used but the 2 in F^2 is also a coefficient.
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2 = (x)(0.1+2x)2 ***This is a common ion effect question, therefore the .1***
 
Both of the examples you gave are correct. The moles of the ion will always be the exponent for these Ksp questions. In the second example you give, there is 2x instead of x because for each molecule of MX2 that dissociates, you get two ions of X-. This is simply from stoichiometry.

What you need to determine varies from question to question. If the question is asking for the Ksp expression then you would simply say Ksp = [Ba2+][F-]^2 like in your first example. If they give you molar solubility, which is x in your second example, and they want you to find the Ksp, then you would plug the value they give for x into 4x^3 to determine Ksp.

Hope this helps!
 
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Bump, still need some help, guys!

Shonster460 is correct. When asked for the Ksp formula, Chad mentions to NEVER include the 2x in the Ksp. However, when you begin to solve, you will need to write it in.

ex. BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2
Ksp = (x)(2x)^2

The only time you do not include the 2x in your calculation is if they give you a common ion or concentration of one of the products. Then they are both x. Hope this helps.
 
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