# $K_{sp}$ value and solubility

1. Apr 30, 2015

### mooncrater

1. The problem statement, all variables and given/known data
I read this line in my textbook:
"$PbS,CdS$ are precipated in dilute solution only due to higher $K_{sp}$"
And I googled about the relation between solubility and $K_{sp}$ and I found that:
"More is the $K_{sp}$ of a salt more it is soluble in a solution(I think water)"
So are the two statements in accordance to each other?If no then which one is wrong?

2. Relevant equations

3. The attempt at a solution
I am confused. I do think that one of these lines should be wrong and I would prefer the first one. But I am not sure about it because of which I have asked it here.

2. Apr 30, 2015

### x86

Like any equilibrium constant, for the reaction aA + bB <===> cC + dD, K = [C]^c[D]^d/ ([A]^a^b).

Therefore, if your reaction is solid <===> solute then a higher Ksp will mean that more solute will dissolve in the solution. (Note that in the equilibrium constant you don't include the solids, so in this case Ksp = [solute])

I'm also confused about the first statement, but I know the above is true

3. Apr 30, 2015

### mooncrater

So, using your statement in this line
I may say that instead of $higher$ it should be $lower$.
What do you think?

4. Apr 30, 2015

### x86

Yes, that makes sense. If Ksp is really low then there will be very low solubility.

5. May 13, 2015

### James Pelezo

You may already know this, but be careful using Ksp as an indication of solubility. You can use Ksp to compare solubility only if the salt ionization ratios are the same. If not, you'll need to use the solubility equation for the ionization ratio. Here's an example where you can get into trouble... Consider single salt saturated aqueous solutions of MgCO3 (Ksp = 1 x 10-5) and BaF2 (Ksp = 1 x 10-6). Is MgCO3 more soluble than BaF2 because it has a larger Ksp-value? NO!

Solubility of MgCO3 = (Ksp)1/2 = (1x10-5)1/2 = 0.0032M b/c MgCO3 has a 1:1 ionization ratio

Solubility of BaF2 = (Ksp/4)1/3 = (1 x 10-6/4)1/3 = 0.0063M b/c BaF2 has a 1:2 ionization ratio

Solubility of BaF2 > Solubility of MgCO3 ... Just an FYI...
Have a good day.