Answer:
0.168mL of the 1.0M KI solution must be added
Step-by-step explanation:
The PbI₂ is in equilibrium with water as follows:
PbI₂ ⇄ Pb²⁺ + 2I⁻
Where Ksp is:
Ksp = 1.4x10⁻⁸ = [Pb²⁺] / [I⁻]²
As molarity of Pb²⁺ is 0.11M:
1.4x10⁻⁸ = [0.11M] / [I⁻]²
1.27x10⁻⁷ = [I⁻]²
3.57x10⁻⁴M = [I⁻]
Thus, to precipitate all Pb²⁺ you need to add 3,57x10⁻⁴M of I⁻. As volume of the solution is 470.0mL = 0.47L, you need to add:
0.47L * (3,57x10⁻⁴moles / L) = 1.68x10⁻⁴ moles of I⁻ = Moles of KI.
That comes from the 1.0M KI. You need to add:
1.68x10⁻⁴ moles of KI * (L / 1.0 mol) = 1.68x10⁻⁴L =
0.168mL of the 1.0M KI solution must be added