Question

Kererences A solution is produced by stirring 1.7000 gram of lead(II) fluoride in 1.00 liter of water at 25°C. Careful analysis shows that 0.515 grams of lead(II) fluoride has dissolved. Calculate the Ksp for lead(II) fluoride based on these data. Кр

Answer 1

s = mass of PbF2 dissolved / volume of solution

= 0.515 g / 1.00 L

= 0.515 g/L

Molar mass of PbF2,

MM = 1*MM(Pb) + 2*MM(F)

= 1*207.2 + 2*19.0

= 245.2 g/mol

Molar mass of PbF2= 245.2 g/mol

s = 0.515 g/L

To covert it to mol/L, divide it by molar mass

s = 0.515 g/L / 245.2 g/mol

s = 2.1*10^-3 mol/L

At equilibrium:

PbF2 <----> Pb2+ + 2 F-

   s 2s

Ksp = [Pb2+][F-]^2

Ksp = (s)*(2s)^2

Ksp = 4(s)^3

Ksp = 4(2.1*10^-3)^3

Ksp = 3.706*10^-8

Answer: 3.71*10^-8