Problem: A solution contains 2.0 X 10 -3 M Ce3+ and 1.0 X 10 -2 M IO3 3-. Will Ce(IO3)3(s) precipitate? [Ksp for Ce(IO3)3 is 3.2 X 10-10.]

• Q > K_sp: the solution is supersaturated and a precipitate will form. Reactants are favored.

• Q = K_sp: the solution is at equilibrium and no precipitate will form.

• Q < K_sp: the solution is unsaturated and no precipitate will form. Products are favored.

For Ce(IO₃)₃(s):

The sulfate ion, IO₃^–, has a charge of –1. Ce then has a charge of +3. 

The dissociation of Ce(IO₃)₃(s) in water is as follows:

Ce(IO₃)₃(s) ⇌ Ce³⁺(aq) + 3 IO₃^–(aq)

Calculate the reaction quotient for Ce(IO₃)₃ is:

$$\begin{gathered} \mathbf{Q}\mathbf{=}\frac{\mathbf{products}}{\cancel{\mathbf{reactants}}} \\ \mathbf{Q}\mathbf{=}\left[{\mathrm{Ce}}^{3+}\right]{\left[{{\mathrm{IO}}_3}^{-}\right]}^{\mathbf{3}} \end{gathered}$$

Note that each concentration is raised by the stoichiometric coefficient: [Ce³⁺] is raised to 1 and [IO₃^–] is raised to 3.

Solving for Q: