### Calculate Ksp given a weight/volume percentage

Example #1: The solubility of lead(II) fluoride is 0.0510%(w/v) at 25 °C. Calculate the Ksp of this salt at this temperature.

Solution:

1) 0.0510%(w/v) means:

0.0510 grams of PbF2 in 100. mL of solution.

2) Determine moles of PbF2 present (the molecular weight of PbF2 is 245.196 g mol¯1):

0.0510 g / 245.196 g mol¯1 = 2.08 x 10¯4 mol
3) Determne the molarity of PbF2:
2.08 x 10¯4 mol / 0.100 L = 2.08 x 10¯3 mol/L.
4) Determine Ksp of PbF2:
PbF2 ⇌ Pb2+ + 2F¯

Ksp = [Pb2+] [F¯]2

Ksp = (2.08 x 10¯3) (4.16 x 10¯3)2

Ksp = 3.60 x 10¯8

Example #2: The solubility of lead(II) fluoride in water is 0.0175%(w/v) at 29 °C. Calculate the Ksp of this salt at this temperature.

Solution:

0.0175%(w/v) = 0.0175 g/100 mL = 0.175 g PbF2 / L

The molar mass of PbF2 is 245.2 g/mol.

(0.175 g / L) / (245.2 g/mol) = 0.0007137 mol/L

PbF2 ⇌ Pb2+ + 2F¯

When PbF2 dissolves:

[Pb2+] = 0.0007137 mol/L

[F¯] = 0.0014274 mol/L

Ksp = [Pb2+][F¯]2

Ksp = (0.0007137) (0.0014274)2

Ksp = 1.45 x 10¯9