Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (Ksp of PbCl2 = 3.2 x 10-8, atomic mass of Pb = 207 u).

To calculate the volume of water required to dissolve 0.1 g0.1g of lead (II) chloride (PbCl2PbCl2) to get a saturated solution, we first need to determine the number of moles of PbCl2PbCl2 dissolved. Then, we can use the solubility product constant (KspKsp) to find the concentration of Pb2+Pb2+ ions in the saturated solution, which will allow us to calculate the volume of water required.

First, let's calculate the number of moles of PbCl2PbCl2 dissolved:

1. Calculate the molar mass of PbCl2PbCl2: Molar mass of Pb=207 g/molMolar mass of Pb=207g/mol Molar mass of Cl=35.453 g/molMolar mass of Cl=35.453g/mol Molar mass of ( \text{PbCl}_2 = \text{Molar mass of Pb} + 2 \times \text{Molar mass of Cl} ] =207+2×35.453=207+70.906=277.906 g/mol=207+2×35.453=207+70.906=277.906g/mol

2. Calculate the number of moles of PbCl2PbCl2: Number of moles=MassMolar mass=0.1 g277.906 g/molNumber of moles=Molar massMass=277.906g/mol0.1g Number of moles≈3.598×10−4 molNumber of moles≈3.598×10−4mol

Now, let's use the solubility product constant (KspKsp) to find the concentration of Pb2+Pb2+ ions in the saturated solution:

PbCl2→Pb2++2Cl−PbCl2→Pb2++2Cl−

Given Ksp=3.2×10−8Ksp=3.2×10−8, we can set up an ice table:

Substituting the equilibrium concentrations into the KspKsp expression:

Ksp=[Pb2+]×[Cl−]2Ksp=[Pb2+]×[Cl−]2

3.2×10−8=(3.598×10−4)×(2×3.598×10−4)23.2×10−8=(3.598×10−4)×(2×3.598×10−4)2

Solve for [Pb2+][Pb2+]:

[Pb2+]=3.2×10−8(2×3.598×10−4)2[Pb2+]=(2×3.598×10−4)23.2×10−8

[Pb2+]≈2.222×10−2 M[Pb2+]≈2.222×10−2M

Now, we can use the concentration to calculate the volume of water required to dissolve 0.1 g0.1g of PbCl2PbCl2:

[Pb2+]=Amount of substanceVolume of solution[Pb2+]=Volume of solutionAmount of substance

Volume of solution=Amount of substance[Pb2+]Volume of solution=[Pb2+]Amount of substance

Volume of solution=3.598×10−4 mol2.222×10−2 MVolume of solution=2.222×10−2M3.598×10−4mol

Volume of solution≈0.0162 LVolume of solution≈0.0162L

Therefore, approximately 0.0162 L0.0162L of water is required to dissolve 0.1 g0.1g of lead (II) chloride to get a saturated solution.