Question

Be sure to answer all parts.

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 210.0 mL of a solution that is 0.13 M in Pb²⁺ion to precipitate all the lead ion?

(b) What mass of PbI₂ should precipitate?

Answer 1

(a) Volume of KI solution = 54.6 mL

(b) Mass PbI₂ = 12.6 g

Explanation

(a) volume of solution = 210.0 mL

molarity Pb²⁺ = 0.13 M

moles Pb²⁺ = (molarity Pb²⁺) * (volume of solution)

moles Pb²⁺ = (0.13 M) * (210.0 mL)

moles Pb²⁺ = 27.3 mmol

moles I^- = 2 * moles Pb²⁺

moles I^- = 2 * (27.3 mmol)

moles I^- = 54.6 mmol

volume KI = (moles I^-) / (molarity KI)

volume KI = (54.6 mmol) / (1.0 M)

volume KI = 54.6 mL

(b) moles Pb²⁺ = 27.3 mmol = 0.0273 mol

moles PbI₂ = moles Pb²⁺

moles PbI₂ = 0.0273 mol

mass PbI₂ = (moles PbI₂) * (molar mass PbI₂)

mass PbI₂ = (0.0273 mol) * (461.0 g/mol)

mass PbI₂ = 12.6 g