To solve this problem, we need to use the concept of solubility product constant (Ksp) and the stoichiometry of the reaction.
First, let's write the chemical equation for the precipitation reaction:
Pb2+ + 2IO3- β Pb(IO3)2(s)
According to the stoichiometry, for every Pb2+ ion, we need two IO3- ions to form one molecule of Pb(IO3)2.
We are given the initial concentrations of Pb2+ and IO3- after mixing the solutions: [Pb2+] = 0.0030 M and [IO3-] = 0.040 M.
Let's assume that x represents the change in concentration of Pb(IO3)2 (which is equal to the solid) at equilibrium. Then, the concentration of Pb2+ and IO3- ions at equilibrium will be (0.0030 - x) M and (0.040 - 2x) M, respectively.
The solubility product expression (Ksp) for Pb(IO3)2 is given as:
Ksp = [Pb2+][IO3-]^2
Substituting the equilibrium concentrations into the solubility product expression, we have:
Ksp = (0.0030 - x)(0.040 - 2x)^2
Now, solve for x by setting the Ksp expression equal to the given Ksp value of 2.5 Γ 10^-13:
(0.0030 - x)(0.040 - 2x)^2 = 2.5 Γ 10^-13
The next step involves solving this equation. However, it is a complicated equation, so I would recommend using a numerical solver or a graphing calculator to find the value of x.
Once you have found the value of x, substitute it back into (0.0030 - x) and (0.040 - 2x) to determine the final concentrations of Pb2+ and IO3- after the precipitation reaction comes to equilibrium.
Remember to check the validity of your answer by ensuring that the equilibrium concentrations are within the given initial concentrations.