Question

A solution of 2.0*10^-4M of silver (I) ions and 1.5*10^-3 M lead(ii) ions is titrated with a sodium iodide solution. Given that the Ksp of silver (I) iodide is 8.3*10^-17 and the Ksp of lead (II) iodide is 7.9*10^-9, which will precipitate first and at what concentration of iodide ions?

Answer 1

   AgI --------> Ag⁺ + I^-

Ksp = [Ag⁺][I-]

8.3*10^-17 = 2*10^-4 * [I-]

[I-]         = 8.3*10^-17/2*10^-4

              = 4.15*10^-13 M

PbI2 -------> Pb⁺² + 2I^-

Ksp    = [Pb⁺²][I-]²

7.9*10^-9 = 1.5*10^-3 *[I-]²

[I-]²         = 7.9*10^-9/1.5*10^-3

[I-]²          = 5.26*10^-6

[I-]           = 2.3*10^-3 M

low conc of [I-] form AgI So AgI will form precipitate first