Question

Can you please explain the process?
When SrF2, strontium fluoride, is added to water, the salt dissolves to a very small extent according to the reaction below. At equilibrium the concentration of Sr2+ is found to be 0.00105 M. What is the value of Ksp for SrF2?
SrF2(s) ⇌ Sr2+(aq) + 2 F-(aq)

Answer 1

An easy one. There are two ways to do this. Actually both are the same but they look different. I'll show you both.
......................SrF2(s) ⇌ Sr2+(aq) + 2 F-(aq)
I.......................solid.........0.................0
C.....................solid..........x................2x
E.....................solid..........x................2x
Ksp = (Sr^2+)(F^-)^2. So you plug in the E line into this Ksp expression to get Ksp = (x)(2x)^2 = 4x^3
Then the problem tells you that at equilibrium (the E line above) that x is 0.00105 M. So 4*(0.00105)^3 = Ksp and you turn the crank.\

The other way is just as simple but I think it looks easier. I think the first approach is easier but here is the second one.
.......................SrF2(s) ⇌ Sr2+(aq) + 2 F-(aq)
Equilibrium......solid.........0.00105M....2*0.00105M
Ksp = (Sr^2+)(F^-)^2 = (0.00105)(0.00210)^2 = ?
Post your work if you get stuck.