Describe how you would use a radioactive iodine isotope to demonstrate that the following process is in dynamic equilibrium: $$\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q)$$

Dynamic equilibrium is a state of balance achieved by a reversible chemical reaction. Even though the reactants continuously transform into products and vice-versa, the concentrations of the involved chemical species stay constant as the rates of the forward and backward reactions are equal.

Radioactive isotopes are often used as tracers in scientific experiments because they emit radiation that can be detected and measured. If a radioactive iodine isotope is introduced to the iodide in this reaction, it can be traced as it moves through the reaction.

To demonstrate dynamic equilibrium, one could begin with a solid precipitate of lead iodide (PbI2), and add a solution that contains the radioactive iodine isotope. In time, the radioactive isotope would be observed in the solid PbI2, proving that not only does the reaction 'forward' direction occur (PbI2 dissolving), but the 'reverse' direction occurs as well (Pb2+ and I- recombining to form PbI2). Measuring the radiation level from the starting point, will allow an increase of radioactivity in PbI2 to show that the dissolved radioactive iodine ions are being re-deposited as solid PbI2. This simultaneous happening of forward and backward reactions, while maintaining a constant level of reactants and products, confirms the dynamic equilibrium state of the reaction.