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⁺.
[tex]100.0 \times 10^{-3} L \times \frac{0.15mol}{L} =0.015mol[/tex]
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:
[tex]\frac{0.030mol}{1.0mol/L} =0.030L=30mL[/tex]
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:
[tex]0.015mol.\frac{461.01g}{mol} =6.9g[/tex]
The mass of PbI2 produced is 6.9 g.
Stoichiometry has to do with the relationshop between mass and mole or mass and volume in a chemical reaction.The reaction equation here is; Pb2+(aq) + 2I-(aq) = PbI2(s).
Number of moles of Pb^2+ ions = 0.15M * 100/1000 L = 0.015 moles
Since 1 mole of Pb^2+ ions reacts with 2 moles of iodide ion
0.015 moles of Pb^2+ ions reacts with x moles of iodide ion
x = 0.015 moles * 2 moles /1 mole = 0.03 moles
Since number of moles = concentration * volume
Volume of iodide solution = 0.03 moles/1.0 M = 0.03L or 30 mL
The mass of PbI2 produced is 0.015 moles * 461 g/mol= 6.9 g
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