Calculate the solubility in g/L of lead (II) fluoride in water at 25 oC if Ksp of PbF2 is 4.1 x 10-8. a. 9.6x10-7 g/L b. 7.4x10-2 g/L c. 4.1x10-2 g/L d. 5.3x10-1 g/L e. 15.6x10-7 g/L

Ibelieve your answer is true 

Well to my expert knowledge and my exquisite calculations i think the the negative ion # 412 is the key to sucess

The density of water is 1 g/cm cubed and in order for the pebble to sink its density has to be greater than the density of water and in order for the twig to float its density has to be less than the density of water

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.