Answer :
Final answer:
The minimum volume of a 0.267M potassium iodide solution required to completely precipitate all of the lead in 135.0ml of a 0.140M lead (II) nitrate solution is 101.2ml.
Explanation:
To determine the minimum volume of the potassium iodide solution required to completely precipitate all of the lead in the lead (II) nitrate solution, we need to consider their stoichiometry and use the concept of limiting reactants. From the balanced equation, we can see that 2 moles of potassium iodide react with 1 mole of lead (II) nitrate to produce 1 mole of lead iodide. Therefore, the mole ratio between the two is 2:1. Using the given concentrations, we can convert the moles to volumes:
Concentration of lead (II) nitrate = 0.140 M
Volume of lead (II) nitrate = 135.0 mL = 0.135 L
Concentration of potassium iodide = 0.267 M
Volume of potassium iodide = x mL
Now, we can set up an equation using the mole ratio:
(0.140 M) * (0.135 L) = (0.267 M) * (2 mol KI / 1 mol Pb(NO3)2) * (0.00135 L)
Solving for x, the minimum volume of the potassium iodide solution required, we get x = 0.1012 L = 101.2 mL.
