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 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?

#### Homework Answers

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

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