Question

A solution contains 1.05E-2 M lead acetate and 8.21E-3 M silver nitrate. Solid sodium iodide is added slowly to this mixture. What is the concentration of silver ion when lead ion begins to precipitate? The Ksp of lead iodide is 8.70E-9 The Ksp of silver iodide is 1.50E-16

Answer 1

initial cocentration of Pb(II) ion is 1.0510^-2 M

and initial concentration of silver ion is 8.21 10^-3 M

now when sodium ion is added it will form Lead Iodide. and silver Iodide

the Ksp of Lead iodide is 8.7 10^-9

then ksp = [Pb²⁺] [I^-]²

or [I^-]² = ksp/[Pb²⁺] =  8.7 10^-9 / 1.0510^-2 = 8.28 10^-7

or,   [I^-] =  8.28 10^-7 = 9.09 10^-4

so when the solubility of lead ion and Iodide ion exceeds the Ksp it willl start to precipitate.

now

ksp of silver Iodide = [Ag][I^-]

1.510^-16 = [Ag⁺] 9.09 10^-4

or,[Ag⁺] =  1.510^-16 / 9.09 10^-4 = 1.65 10^-13

so thhis amount of silver ion is insoluble in the solution and exists as Silver Iodide

hence concentration of silver ion present is

[Ag⁺] = 8.2110^-3 - 1.65 10^-13 = 8.2010^-3 M