You have 310. mL of a 5.75×10-5 M Pb(NO3)2 solution. How many moles of NaCl must be added to precipitate a solid?
The Ksp of PbCl2 is 1.6×10-5.
I'm not sure where to go after I make the ICE table
After creating the ICE table, you need to use the solubility product constant (Ksp) - in this case, for PbCl2 - to determine the concentration of Pb2+ and Cl- ions in the solution.
The balanced equation for the dissociation of PbCl2 is: PbCl2 ⇌ Pb2+ + 2Cl-
Let's assume x represents the concentration of Pb2+ and 2x represents the concentration of Cl-. Since the concentration of Pb2+ is equal to the concentration of Cl- (according to the stoichiometry of the balanced equation), we can simplify it as follows:
[Pb2+] = x
[Cl-] = 2x
From the information given, you have a total volume of 310 mL of the Pb(NO3)2 solution, with a concentration of 5.75×10-5 M. To find the initial concentration of Pb2+ ions (x), you need to convert the volume to liters:
Volume of solution (L) = 310 mL = 0.310 L
Now you can use the definition of concentration to find the initial concentration of Pb2+ ions:
[Pb2+]initial = concentration = 5.75×10-5 M
Substituting this value for [Pb2+]initial in the ICE table, we have:
Initially (I): [Pb2+] = 5.75×10-5 M, [Cl-] = 2(5.75×10-5) M = 1.15×10-4 M
Change (C): -x, -2x
Equilibrium (E): [Pb2+] = 5.75×10-5 - x M, [Cl-] = 1.15×10-4 - 2x M
Now that you have the expressions for the concentrations of Pb2+ and Cl- at equilibrium, you can substitute them into the expression for Ksp:
Ksp = [Pb2+][Cl-]^2 = (5.75×10-5 - x)(1.15×10-4 - 2x)^2
Since you're interested in finding the number of moles of NaCl needed to precipitate a solid, you can use stoichiometry to determine the mole ratio between PbCl2 and NaCl. The ratio is 1:2, meaning for every 1 mole of PbCl2 precipitated, 2 moles of NaCl are needed.
To find the number of moles of PbCl2 precipitated, you need to set up an equation with the Ksp expression equal to the value of Ksp:
Ksp = [Pb2+][Cl-]^2 = (5.75×10-5 - x)(1.15×10-4 - 2x)^2 = 1.6×10-5
Solving this equation will give you the value of x, which represents the concentration of Pb2+ ions at equilibrium. Once you have x, multiply it by 0.310 L to find the number of moles of PbCl2 precipitated. Finally, multiply that result by 2 to get the number of moles of NaCl needed to precipitate the same amount of PbCl2.