A solution is prepared by mixing 57.8 mL of 0.24 M Pb(NO3)2 with 58.6 mL of 1.4 M KCl. Calculate the concentrations of Pb2+ and Cl - at equilibrium. Ksp for PbCl2(s) is 1.6 ✕ 10-5.
To solve this problem, we need to use the concept of the solubility product constant (Ksp) to determine the concentrations of Pb2+ and Cl- at equilibrium.
1. First, let's calculate the moles of Pb(NO3)2 and KCl:
Moles of Pb(NO3)2 = volume (in L) × concentration = 57.8 mL × 0.24 mol/L = 0.0139 mol
Moles of KCl = volume (in L) × concentration = 58.6 mL × 1.4 mol/L = 0.082 mol
2. Next, let's calculate the possible formation of PbCl2 based on the stoichiometry:
Pb(NO3)2(aq) + 2KCl(aq) → PbCl2(s) + 2KNO3(aq)
Since 1 mole of Pb(NO3)2 reacts to form 1 mole of PbCl2, the maximum number of moles of PbCl2 that can form is 0.0139 mol.
3. Now, let's calculate the concentrations of Pb2+ and Cl- at equilibrium:
Since all of the Pb(NO3)2 will react to form PbCl2, and moles of Pb(NO3)2 = moles of Pb2+ at equilibrium, the concentration of Pb2+ is equal to the initial concentration:
[Pb2+] = 0.24 M
Since 1 mole of PbCl2 is formed for each mole of Pb(NO3)2, the concentration of Cl- at equilibrium is:
[Cl-] = moles of PbCl2 formed / total volume (in L) = 0.0139 mol / (58.6 mL + 57.8 mL) = 0.0139 mol / 0.1164 L = 0.1193 M
Note: The concentrations have been rounded to the appropriate number of significant figures.
Therefore, the concentrations of Pb2+ and Cl- at equilibrium are approximately 0.24 M and 0.1193 M, respectively.