Question

How do you write Ksp?

Answer 1

It is asking for the steps in order to determine Ksp.

Answer 2

You would write it like you would with any other equilibrium expression, the difference is that the "reactant" is a solid, so it's concentration is equal to one - so it is omitted.

You would use an I.C.E. table and substitute into the expression.

Do you have a question we can do as an example?

Answer 3

BaCrO(s)<-->Ba^2+ +CrO4^2-

Answer 4

I think you typed that incorrectly, you're missing the subscript "4". It's balanced so we can proceed. Do you know how to write an equilbrium expression? like,

\(\sf K=\dfrac{[products]}{[rectants]}\)

Answer 5

That is supposed to be BaCro4(s) I apologize. The other side are both (aq)

Answer 6

I do know how to write products over reactants but I am not sure of anything after that point.

Answer 7

Okay cool, well what's important is that you identify what the products and reactants are and their respective stoichiometric coefficients (which in this case are all 1)

\(K=\sf \dfrac{[Ba^{2+}]^1 [CrO4^{2-}]^1}{[BaCrO_4]^1}\)


Since \(\sf BaCrO_4\) is a solid, it's concentration is constant and thus we write 1, so we can omit it because it doesn't change the numerator. We can rewrite and now call it "the solubility product constant expression.

\(K_{sp}=\sf [Ba^{2+}] [CrO4^{2-}]\)

Answer 8

Ok. For other things though, like for instance, it is unbalanced, how would I know how to write Ksp?

Answer 9

Now say you wanted to find the concentration of these ions given the Ksp.

We need to write an I.C.E. table, where I is the initial concentration, C is the change and E is the concentration at equilibrium (once it's dissolved as much as it can).

\(\sf [Ba^{2+}] ~~~~~[CrO4^{2-}]\)
I 0 0 (nothing has yet dissolved)

C +x +x (x amount dissolved)

E x x

\(K_{sp}=\sf [Ba^{2+}] [CrO4^{2-}]=x*x=x^2\)

Answer 10

Lets do another example where all the coefficients aren't 1

Answer 11

I don't understand the ICE table...?

Answer 12

What don't you understand?

Answer 13

It is to find the ions but I don't see how it does exactly. Considering you have only the element abbreviation, how do you end up with the ions?

Answer 14

so if you notice at E, the concentrations at equilibrium are marked as "x"

once you find what "x" is

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in the case above, given the Ksp, you would solve like this:

\(Ksp=x^2\rightarrow x=\sqrt{Ksp}\)

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you would use the x value as the concentration of each ion. This would be different if the coefficients weren't 1.

this video is good because you can see the process in which he writes it. watch it and ask me if you have doubts.

https://www.youtube.com/watch?v=tUnbm379Wfk

Answer 15

Ok. I will give it a watch. Can you help with one more problem?
At 25degreesC, a saturated solution of cerium(III) hydroxide (Ce(OH)3) contains 5.1x10^-6 moles of the compound in one liter of solution. What is the reaction equation?

Answer 16

The "dissolving" (called "dissolution") reaction?

\(\sf Ce(OH)_3\rightleftharpoons Ce^{3+}+OH^- \)

the charge on the cerium ion is known because the polyatomic ion hydroxide (\(OH^-\)) is negative -1 and there are 3 of these ions.

The charges must be balanced (the net charge once these are added should equal to 0)

Answer 17

I forgot a coefficient: \(\sf Ce(OH)_3\rightleftharpoons Ce^{3+}+3~OH^-\)