Question

328. mL of 0.00345 M NaI (aq) is combined with 703. mL of 0.00802 M Pb(NO3)2 (aq). Determine if a precipitate will form given that the Ksp of Pbl2 is 1.40x10-8.

Answer 1

Answer:

The solution will not form a precipitate.

Explanation:

The Ksp of PbI₂ is:

PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)

Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] Concentrations in equilibrium

When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:

[I⁻] = 0.00345M × (328mL / (328mL+703mL) = 1.098x10⁻³M

[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) = 5.469x10⁻³M

Q = [I⁻]²[Pb²⁺] Concentrations not necessary in equilibrium

If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.

Replacing:

Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹

As Q < Ksp, the solution is not saturated and will not form a precipitate.