Question

One way to remove lead ion from water is to add a source of iodide ion so that lead iodide will precipitate out of solution: Pb2+(aq) + 2I-(aq) = PbI2(s).
a. What volume of a 1.0 M KI solution must be added to 100.0 mL of a solution that is 0.15M in Pb2+ ion to precipitate all the lead ion?
b. What mass of PbI2 should precipitate?

Answer 1

Answer:

a. 30 mL

b. 6.9 g

Explanation:

Let's consider the following precipitation reaction.

Pb²⁺(aq) + 2I⁻(aq) ⇄ PbI₂(s)

a. What volume of a 1.0 M KI solution must be added to 100.0 mL of a solution that is 0.15M in Pb²⁺ ion to precipitate all the lead ion?

First, we will calculate the moles of Pb²⁺.

$100.0 \times 10^{-3} L \times \frac{0.15mol}{L} =0.015mol$

The molar ratio of Pb²⁺ to I⁻ is 1:2. Then, we have 2 × 0.015 mol = 0.030 mol of I⁻. There is 1 mole of I⁻ per mole of KI, so we also have 0.030 mol of KI. The volume of the 1.0 M KI solution is:

$\frac{0.030mol}{1.0mol/L} =0.030L=30mL$

b. What mass of PbI₂ should precipitate?

The molar ratio of Pb²⁺ to PbI₂ is 1:1. Then, we have 0.015 mol of PbI₂. The molar mass of PbI₂ is 461.01 g/mol, and the mass corresponding to 0.015 moles is:

$0.015mol.\frac{461.01g}{mol} =6.9g$