[lizardo5]

if the ksp=6.17E-11 for MX2.  What are the ion concentrations in a saturated solution of MX2?

[M^2+]
and [X^1-]

TIA

[Sam (NG)]

Please read The Forum Rules. In particular, take note of:

4.) Please show that you've at least attempted the problem.  We don't mind helping you solve problems but we are ethically opposed to doing homework for you. Violators will have their topic deleted or locked.

[lizardo5]

I am unsure of how to even attempt this problem.  Can someone please get me started in the right direction?
tia

[enahs]

http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Solubility_Products.htm

[lizardo5]

I am given Ksp though ... not a certain concentration?

[Sam (NG)]

I am given Ksp though ... not a certain concentration?

Is This not what you are trying to figure out?

[enahs]

I am given Ksp though ... not a certain concentration?

Is This not what you are trying to figure out?

Why yes, yes it is. No wonder why I gave him that link, eh? 

[lizardo5]

i don't understand how to go about using the website b/e all of those use formulas that are given.  Mine does not have a formula given it is using MX

[enahs]

From this site:

Since the equilibrium constant refers to the product of the concentration of the ions that are present in a saturated solution of an ionic compound, it is given the name solubility product constant, and given the symbol Ksp.

M is species 1, and X is species 2.
X is squared because in the chemical equation you have

MX₂ = M + 2 X

You know this, because that is how the solubility product constant is defined. You can also look at the charges on the ions, and notice that to balance the charges (make the original compound neutral) you would need 2 of whatever X happens to be.


Your question is exactly the same as the second question on that site, just with a different KsP, only your species are called X and M

[lizardo5]

ok. so i created my own formula:
M^2+ + X^1- -> MX2

From here i said that the ksp = 6.17E-11 =

• [x^2]

I solved for x and got x= .000395

Then i said that
[M^2+]=.000395 and
[X^1-] = 1.56025e-7

any more suggestions or am i all wrong?  

[enahs]

You are wrong.
I see the time stamp on your post was like 20 seconds after my other one; go back and read it in case you missed it.

[english]

This is a simple equation.  Solve for x.

Ksp = [X²⁺][Y ^-]²

or

Ksp = x³

[enahs]

This is a simple equation.  Solve for x.

Ksp = [X²⁺][Y ^-]²

or

Ksp = x³

No, it simplifies down to 4x³. But the point is to not just tell him an equation, but to show him how to figure it out on his own.

[english]

I don't see where the 4 is coming from.  Did I miss something?


Always write an equation first.  Any problem can be solved with an equation to look at.

XY_2(s) <----> X²⁺_(aq) + 2Y ^-_(aq)

lizardo Ksp is much like Kw, in that the solid mass (precipitate) is constant.  Treated as a pure solid, we have

K = [X^2+][Y ^-]²
        [XY₂]

Since XY₂ is constant, we carry this term to K, giving

K[XY₂] = [X²⁺][Y ^-]²

or alternatively,

Ksp = [X²⁺][Y ^-]², where if the concentrations of both ions are unknown, they can be designated as x.  Notice that the anion is x² because there are 2 mol of anion for every 1 mol of cation.

This can be written as Ksp = x*x², or Ksp = x³

[enahs]

No, not quite.

Because you have a 2 in front of the Y species in the balanced chemical equation, it means you have twice as much Y as you do X. So to represent the quantity of Y as a function of X, Y= 2*X.
Because the solubility product constant is defined as the concentrations raised to their stoichiometric coefficient, you are left with:

This leaves you with Ksp = X * (2X)² = 4X³


If you do it your way,  say (for this case)
Ksp = X * X², when you solve for X it will be off.

example

Ksp = 10; AB₂    <---> A + 2 B              (equation 1)
Your method:
Ksp = [A]² = [X][X]²           (equation 2, as you are writing)
10 = X³
X= 2.15
Plugging back into equation 2, you are then saying the concentration of A is 2.15 and Y is 2.15, this can not be so (remember, the power of 2 in equation 2 is only there because how we define the solubility product constant)

now, I say (which is correct):
Ksp = 10; AB₂      <----->    A   + 2 B            (equation 3)
Ksp = [A][2*B]²   =   [X][2*X]²       (equation 4)
10 = 4X³
X = 1.357

Plugging back into equation 4 we can now get A = 1.357 and B = 2.71


Again, I refer to the same site:
http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Solubility_Products.htm