posted by Anmol .
The Ksp for PbF2 is 7.1 x 10 to the(-7). The Kb for the F- anion is 1.4 x 10 to the(-11). A student weighs out 2.5 g of PbF2, places it into 100mL of deionized water and stirs.
a) Will there be any solid PbF2, left in the flask?
b) What is the pH of the solution?
PbF2 ==> Pb^+2 + 2F^-
Ksp = (Pb^+2)(F^-)
Set up an ICE chart and solve for solubility PbF2 and for F^-.
Use F^- in the equation below for pH.
F^- + HOH ==> HF + OH^-
Kb = (HF)(OH^-)/(F^-).
Set up ICE chart and solve for OH^-, convert to pOH, then to pH.
For part a), from the solubility of PbF2, (which is in M or moles/L), take a tenth to determine the amount that will dissolve in 100 mL, and compare with 2.5 grams.