Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to –8 get a saturated solution. (Ksp of PbCl2 = 3.2 × 10 , atomic mass of Pb = 207 u).
Show answer
Ksp of PbCl2 = 3.2 × 10 Let S be the solubility of PbCl2. PbCl2 (s) Pb 2+ – (aq) + 2Cl (aq) Concentration of species at t = 0 1 0 0 Concentration of various species at equilibrium 1–S S 2S 2+ – 2 2 3 Ksp = [Pb ] [Cl ] = (S) (2S) = 4S Ksp = 4S –8 K sp 3.2 × 10 –1 –9 –1 S = = mol L = 8 × 10 mol L 4 4 –1 –3 –1 S = 3 8 × 10 –9 = 2 × 10 –3 mol L ∴ S = 2 × 10 mol L 101 Equilibrium Molar mass of PbCl2 = 278 –3 –1 ∴ Solubility of PbCl2 in g L–1 = 2 × 10 × 278 g L –3 –1 = 556 × 10 g L –1 = 0.556 g L To get saturated solution, 0.556 g of PbCl2 is dissolved in 1 L water. 0.1 0.1 g PbCl2 is dissolved in L = 0.1798 L water. 0.556 To make a saturated solution, dissolution of 0.1 g PbCl2 in 0.1798 L ≈ 0.2 L of water will be required. V V V V