Answer:
a. 30 mL
b. 6.9 g
Explanation:
Let's consider the following precipitation reaction.
Pb²⁺(aq) + 2I⁻(aq) ⇄ PbI₂(s)
<em>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?</em>
First, we will calculate the moles of Pb²⁺.
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:
<em>b. What mass of PbI₂ should precipitate?</em>
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: