Common-Ion Effect?

Question

Does the common-ion effect change the value of Ksp? Explain.

Does the common-ion effect change the value of the ion product (Q)? Explain.

please help mee. i can't find it in the bookk

Answer 1

Hi. I'll try to tackle this one. The "common" ion is just that, an ion (say, Cl-) that is common to two different substances in solution. Say you have Silver Chloride, AgCl, that dissociates in water to give:

AgCl--->Ag+ Cl-

And you have a solubility constant, K(sp)=

1.7 X 10-10 = [Ag+] [Cl-]

(The brackets enclose concentrations expressed in Molarity, M)

So it tells us that the K(sp) equals the product of the molar concentrations of each ion. Keep in mind one thing here: If a certain compound produces more than one ion, you must incorporate that into the equation for K(sp).

Say you have A(2)B--->2A+ + B--

Notice above that for each A(2)B, 2 moles of A are produced. We MUST incorporate this into K(sp):

K(sp)= [2A+]^2[B--]

Notice we premultiply A by 2, and then square it, because A yields two ions in water, not just one. Keep this in mind if you get substances that produce more than one ion in solution.

Now say we add 2M regular table salt, NaCl, which itself produces:

Nacl-->Na+ Cl-

The Chloride ion is the common ion here. We expect that by Le Chatelier's principle, more of the Cl- on the right side of
AgCl--->Ag+ Cl-

will make the equation go "backwards". (it will reduce the solubility of AgCl. Do you see why? The idea is that if you add to one side, it will "push" towards the other (more AgCl) and if you subtract, it will "pull" the reaction towards it, more Ag+ and Cl-)

So we know there will be less AgCl in aqueous solution. But how much?

Let's go back to the K(sp) of AgCl:

1.7 X 10-10 = [Ag+] [Cl-]

Since we aded 2M of NaCl, that will give us 2M of common ion Cl-. So we simply add that to the K(sp) equation:

1.7 X 10-10 = [x] [x+2]
1.7 X 10-10 = x^2+2x

The solution for x here is a bit tricky, but we can make a simplifying assumption: because "x" is so small compared to 2M, we assume that:

1.7 X 10-10 = [x][2]= 2x

So x=8.5x10-11M

Compare this value to
1.7 X 10-10 = [x] [x]=x^2
So Square root of 1.7 X 10-10= 1.30 x10-5M

We have reduced the solubility of AgCl drammatically by adding the common ion, from 1.30 x10-5M to 8.5x10-11M.

Notice that the K(sp) does NOT change, it remains the same. What does change is the solubility of the compound when a common ion is added. When a common ion is added, the rule is that the solubility is reduced.

Now as to Q(sp), the ion product. Since we added additional C- from NaCl, the product of [Ag+] * [Cl-]
will be greater than it is in equilibrium, which is the K(sp)
So Q(sp)>K(sp) initially, because there is more of the common ion, Cl-. That is:

[x][x+2]>[x][x]

But the reaction will eventually reach equilibrium, at K(sp), meaning less AgCl dissolves to reach equilibrium.

In equilibrium, of course:
Q(sp)=K(sp).

So the K(sp) stays the same, but initially the ion product increases so that it is greater than the initial equilibrium, because we added more of the common ion.

Answer 2

Common Ion