Question

8. Determine the mass (in g) of PbCl2 (Ksp, PbCl2 = 1.7x10^-5) that will dissolve in 0.450L of 0.375M Pb(NO3)2 solution.

Answer 1

Given data: Ksp of Pb Cl₂ = 1.78105 Volume of Pb (NO2) = 0.450L Concentration of Pb(NO3)₂ = 0.375m Solution Leat the Pb Cola solubility of PbClg be S, then - Pb2+ + 2 at S 25 Pb(NO3)2 is completely ionised because it is strong electrolyte. Pb(NO3)₂ will provide Pb²+ ion whose concentration is 0.375M » [Pbd+] = (5+0.375) [cal-] = 25 Ksp = [Pb2+] [61]? 1:7x105 = (5 +0.375) (23)2 1.7x10-5 = (5 +0.375) (4 S') 1-7X 10-5 = S(5+ 0.375)

4.25 8106 = 3 + 0.375 g : > 3 sin omall quantity and Sy is also very small ( so six neglecting 4.25 x156 = 0.375 52 sa = 4.25*10-6 0.375 8 = 11-33x106 s = 11.33x106 S = 3-366X16"? mal

Solubility of Pb Cl in g/L = Sx molar mass of PbCl2 = 3.366 x 103 malu mal x 270-12 mol = 936-00 X1o 3 g mass of PbCl2 = Solubility of Pb Cla in g x Volume = 936.08 x10g x 0.450 L = 421.24 x 10°g 2 0.42195 Hence mass of pbelg is 421. 24X103