What is the equilibrium concentration?

Question

350. mL of 3.2 M Pb(NO₃)₂ and 200. mL of 0.020 M NaCl are added together. The Ksp for lead(II) chloride is 1.6 x 10⁻⁵. What is the equilibrium concentration of the chloride ion when all PbCl₂(s) has formed?

Answer 1

After mixing, the final concentrations are as follows:

Pb(NO3)₂(aq) + 2NaCl(aq) ==> PbCl₂(s) + 2NaNO₃(aq) ... balanced equation

moles Pb(NO₃)₂ = 0.350 L x 3.2 mol / L = 1.12 moles

moles NaCl = 0.200 L x 0.020 mol / L = 0.004 mols

NaCl is limiting

moles PbCl₂ formed = 0.004 mols NaCl x 1 mol PbCl₂ / 2 mol NaCl = 0.00200 mol PbCl₂

Total final volume = 350 ml + 200 ml = 550 ml = 0.550 L

[PbCl₂] = 0.00200 mols / 0.550 L = 0.00364 M

PbCl₂(s) <==> Pb²⁺(aq) + 2Cl^-(aq)

Q = [Pb²⁺][Cl^-]²

Q = (0.00364)(0.00727)² = 1.9x10^-7

Since Q is < Ksp, a precipitate will not form.

The [Cl^-] will be 0.00727 M