...........PbF2 ==> Pb^2+ + 2F^-
I..........solid...........0.............0
C........solid............x.............2x
E........solid............x..............2x
Ksp = (Pb^2+)(F^-)^2
Ksp = (x)(2x)^2
The problem tells you solubility is 0.830 g/L. How many mols/L is that. Substitute that into the last equation and solve for Ksp.
The solubility of PbF2 in water is 0.830 g/L. What is the value of Ksp?
how would I calculate this?
1 answer