Difference in solubility products and molar solubility?

Question

amtran_bus · Answer

Here are example problems with answers to explain my question. 

1. Calculate solubility product: 
(a) SrF2, 7.3 x 10-2 g/L Ans: 7.8 x 10-10 
Because: 7.3x10-2 / molar mass (125.62) =.000581. Since [F]2, then 4x3. Therefore, 4(.000581)3. 
____ 

2. Determine molar solubility: 
(c) in 0.20 M Pb(NO3)2 solution. Ans: 3.3 x 10-3 M 
Because: Pb is the common ion (that is, it is mentioned in both), then 8.9x10-6=4x2 [.20M], or x=sqrt(8.9x10-6/(4*.20)). 


Why do I have to use different methods to solve? How does solubility and molar solubility differ, and what is their relationship to Ksp? 

Thanks!

amtran_bus · Answer

@AccessDenied @aaronq @abb0t @Hero @hartnn @ganeshie8 @cwrw238 
Thank you for your time, the reason I tagged you is because you are on my fan list.

amtran_bus · Answer

@chmvijay

aaronq · Answer

The molar solubility is just the number of moles that is soluble in 1 L (i.e. molarity).
 
The solubility product (constant <- often omitted) is the product (from the multiplication) of the molar solubilities. This is the Ksp.

You're solving for 2 different things, in your first example you have:  $Ksp=[Sr][F]^2$

if you're solving for the $\color{red}{molar~ solubility}$ then you have to be given the Ksp, and you'd be solving for x.

  $Ksp=4x^3 \rightarrow \color{red}x= (\frac{Ksp}{4})^{1/3}$


if you're solving for the $\color{blue}{solubility~product~constant}$ then you'll be given the molar solubility (in one form of another) and you'll be solving for Ksp.

   $\color{blue}{Ksp}=4x^3 $

amtran_bus · Answer

Thank you aaronq. You probably just saved me 15 points on my next test.

aaronq · Answer

no problem, glad you got that doubt cleared up.