Selective Precipitation
The concentration of iodide ion that will give a saturated solution
of copper(I) iodide is 1.0 x 10-11 M. This concentration will not precipitate the copper(I) ion. It will remove most of the silver ion. Calculating the quantity of silver ion remaining in solution we get:
[Ag+] = = = 8.3 x 10-6 M
Ksp
[I -]
8.3 x 10-17
1.0 x 10-11
Since the initial silver ion was 0.10 M, most of it has been removed,
and essentially none of the copper(I) was removed, so the separation
was quite complete. If the iodide was added as sodium iodide, you
would have to add only a few nanograms of NaI to remove nearly all
of the silver from solution:
1.0 x 10-11 mol I - x x = 1.50 ng NaI
149.9 g NaI
mol NaI
1 molNaI
mol I -
| Previous slide | Next slide | Back to first slide | View graphic version |