A 250.0mL sample of 0.0012M Pb(No3)2 (aq) is mixed with 150.0mL of 0.0640M Na (aq) should precipitation of PbI2 (s) ,Ksp=7.1¡Á10^-9, occur?
To determine if precipitation of PbI2 (s) will occur when the two solutions are mixed, we need to compare the concentration of Pb2+ and I- ions to the solubility product constant (Ksp).
First, let's calculate the moles of Pb2+ and I- ions present in each solution:
For the Pb(No3)2 solution:
Volume = 250.0 mL = 0.250 L
Molarity (M) = 0.0012 M
Moles of Pb2+ ions = Molarity x Volume = 0.0012 mol/L x 0.250 L = 0.0003 mol
For the NaI solution:
Volume = 150.0 mL = 0.150 L
Molarity (M) = 0.0640 M
Moles of I- ions = Molarity x Volume = 0.0640 mol/L x 0.150 L = 0.0096 mol
Now, let's compare the concentrations of Pb2+ and I- ions:
[Pb2+] = 0.0003 mol/0.400 L = 0.00075 M
[I-] = 0.0096 mol/0.400 L = 0.024 M
Now, we can calculate the Qsp (the ion product) using the equation: Qsp = [Pb2+][I-]
Qsp = (0.00075 M)(0.024 M) = 1.8 x 10^-5
Comparing Qsp (1.8 x 10^-5) to Ksp (7.1 x 10^-9), we can see that Qsp is greater than Ksp. This indicates that the solution is already supersaturated with respect to PbI2, and precipitation of PbI2 (s) will occur when the two solutions are mixed.