# Mixing a Tagged Sample of Lead Iodide with another Iodide Salt [closed]

A sample of sparingly soluble $\ce{PbI_2{(s)}}$ containing radioactive $\ce{^{133}I}$ is added to 0.10 M $\ce{KI_{(aq)}}$ and stirred overnight, what happens?

I The radioactivity of the liquid phase increases significantly.

II The concentration of the $\ce{I^{–}}$ ion in solution increases significantly.

The answer is I. I thought II would be true because some of the water evaporates, but I guess not to a "significant" extent. So that marks II off, and leaves II. I'm fairly new to nuclear chem so I don't know much about how/why radioactivity increases, could someone please explain this to me? How could radioactivity "increase?"

## closed as off-topic by Jon Custer, Todd Minehardt, Klaus-Dieter Warzecha, ringo, FreddyApr 19 '16 at 5:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

You are missing a basic aspect of how a precipitate in solution behaves. The precipitate isn't just statically immersed in the solution but rather some is dissolving and some of the ions (or molecules) in solution are precipitating out. So for a salt like $\ce{PbI2}$ there will be an exchange of ions which are in the precipitate with those which are in solution.