Question

Calculate solubility in the presence of a common ion. Calculate the solubility of PbF2 (a) in pure water and (b) in a solution in which [F"] = 0.225 M. M Solubility in pure water = Solubility in 0.225 M F = M

Answer 1

Let us consider a sparingly soluble salt say Ad By. When this salt is dissolved in water to get a saturated solution at a particular temperature, The following equilibrium is established 1 Ax Byl) wAt + y B*] The solubility product for the equilibrium is given ase Ksp = [A 4+][8*-78 where Af and B&- denolē positive and negative ion respectively and a, y represent the no. of these con in the felmula of the electrolyte Let us suppose that the solubility of a sparingly soluble salt An By is smoles/ it, we have, Ax By ř a AYT + y BX- Smoles a Smoles y Smoles [at] = as ; [B7] a gs. Ksp = [43"] * [ex]} = (x5)*(gs)* Scanned with CamScanner na n o

from literalura at 25°C Ksp = 471018 boe het nooit mean lonie PbEz & Pb receptez hobi bilely . mello Smells lit. S es 1. Ksp = [pb*') [f] kup 5xf252 ky - 5x4² kap = 433 Cs Scanned with 6) = 2.et hy molf' CamScanner

D) 9n presence of f = 0.925 14. der be the solubility be smoth it. Pb. f & Pb2+ + 2F S . 3 28 25 +0,225, D-PAS LE INow = (25+ 0.02s) kap = [244] [4] 4x10n8 , sx (25+0.225) Axpone sx (.223) 2 LES : 4x10-es 10.225 V 70.225] 2 = 17.90 X 70-7 modello ES Scanned with CamScanner