How do you know that all the Pb(NO3)2 is used up at equilibrium?

Question

The whole question:

A saturated solution of lead iodate in pure water has a lead ion concentration of 4.0 * 10^-5 mole per liter at 20 degrees C. To 333 milliliters of a 0.120 molar Pb(NO3)2 solution, 667 milliliters of 0.435 molar KIO3 is added. Calculate the concentrations of Pb+2 and IO3- in the solution at equilibrium.

Answer 1

Pb(NO₃)₂(aq) + 2KIO₃(aq) ==> Pb(IO₃)₂(s) + 2KNO₃(aq) ... balanced equation

moles Pb(NO₃)₂ present = 0.333 L x 0.120 mol/L = 0.03996 moles initially present

moles KIO₃ present = 0.667 L x 0.435 mol/L = 0.2901 moles initially present

Limiting reactant is Pb(NO₃)₂ as seen by the 1:2 mol ratio to KIO₃ so it will run out first & KIO₃ is in excess

From the solubility data, we can find the Ksp:

Pb(IO₃)₂(s) <==> Pb²⁺(aq) + 2IO₃^-(aq)

Ksp = [Pb²⁺][IO₃^-]² = (4x10^-5)(4x10^-5)²

Ksp = 2.6x10^-13

After precipitation occurs, we have...

The ppt reaction is Pb²⁺ + 2IO₃^- ==> Pb(IO₃)₂

After ppt reaction...

moles IO₃^- = 0.29 mol - 0.08 mol = 0.21 mol IO₃^- = 0.21 M IO₃^-

And to find moles Pb²⁺ we have...

2.6x10^-13 = [Pb²⁺][0.21]²

[Pb²⁺] = 5.9x10^-12 M Pb²⁺