In an experiment to determine the solubility of lead chloride (PbCl2), 5.6g of (NH4)2SO4 was added to a
250ml solution containing an unknown amount of dissolved lead chloride (PbCl2) resulting in the formation of
Lead sulfate precipitate. The mass of the lead sulfate precipitate was found to be 1.645g. From the
information provided, calculate the mass of PbCl2 that dissolved in the given volume of solvent (250 ml),
hence the solubility and the solubility product constant (Ksp)
11.676g pbcl2 and ksp= 0.642
g PbSO4 = 1.645
mols PbSO4 = 1.645/molar mass = estimated 0.00542 = estd 0.00542mols Pb = estd 0.00542 mols PbCl2.
g PbCl2 = mols PbCl2 x molar mass PbCl2 and that is g/250mL. That x 4 = g/L
Then g/L divided by molar mass PbCl2 = M = (PbCl2) = (Pb^2+).
(Cl^-) = 2 x (Pb^2+).
Then Ksp = (Pb^2+)(Cl^-)^2 = ?
85g
Please the answers
To calculate the mass of PbCl2 that dissolved in the given volume of solvent (250 ml), we need to use the information about the formation of the Lead sulfate precipitate.
1. First, we need to find the moles of (NH4)2SO4 used:
- The molar mass of (NH4)2SO4 is 132.14 g/mol.
- We divide the given mass of (NH4)2SO4 (5.6g) by its molar mass to find the moles:
5.6g / 132.14 g/mol = 0.0424 mol
2. From the balanced chemical equation of the reaction:
PbCl2 + (NH4)2SO4 → PbSO4 + 2NH4Cl
We can see that the reaction has a 1:1 stoichiometric ratio between PbCl2 and PbSO4. This means that for every mole of PbSO4 formed, one mole of PbCl2 dissolved.
So, the moles of PbSO4 (formed by the precipitation) is also equal to the moles of PbCl2 dissolved in the solvent.
3. Next, we find the moles of PbSO4 formed:
- The molar mass of PbSO4 is 303.26 g/mol.
- We divide the mass of the PbSO4 precipitate (1.645g) by its molar mass to find the moles:
1.645g / 303.26 g/mol = 0.00542 mol
4. Since the moles of PbSO4 are equal to the moles of PbCl2 dissolved, we can now calculate the moles of PbCl2 dissolved:
Moles of PbCl2 = 0.00542 mol
5. Finally, we can calculate the mass of PbCl2 dissolved in the given volume of solvent (250 ml):
- We convert the volume of the solvent to liters by dividing by 1000:
250 ml / 1000 = 0.25 L
- Now, we can use the formula: moles = concentration × volume
Concentration = moles / volume
Concentration = 0.00542 mol / 0.25 L = 0.0217 mol/L
To find the mass, we multiply the concentration by the molar mass of PbCl2:
Mass of PbCl2 = Concentration × molar mass
Mass of PbCl2 = 0.0217 mol/L × 278.1 g/mol (molar mass of PbCl2)
Mass of PbCl2 = 6.03 g
Therefore, the mass of PbCl2 that dissolved in the given volume of solvent (250 ml) is 6.03 grams.
To calculate the solubility, we divide the mass of PbCl2 dissolved by the volume of the solvent:
Solubility = Mass of PbCl2 / Volume of solvent
Solubility = 6.03 g / 250 ml
Solubility = 0.0241 g/ml
Lastly, the solubility product constant (Ksp) represents the equilibrium constant for the dissolution of the salt. In this case, it is the product of the concentrations of the ions in the saturated solution of PbCl2. Since we have determined the solubility of PbCl2, we can use that information to calculate Ksp.
For the reaction PbCl2 ⇌ Pb2+ + 2Cl-,
Ksp = [Pb2+][Cl-]^2
Knowing that the solubility of PbCl2 is 0.0241 g/ml, we can convert it to molarity (M):
Molarity = (0.0241 g/ml) / (278.1 g/mol)
Molarity = 8.67 × 10^-5 mol/L
Since the stoichiometric ratio is 1:2 between PbCl2 and Cl-, the concentration of Cl- ions in the saturated solution is 2 times the molarity:
Concentration of Cl- ions = 2 × Molarity
Concentration of Cl- ions = 2 × 8.67 × 10^-5 mol/L = 1.73 × 10^-4 mol/L
Therefore, the solubility product constant (Ksp) for PbCl2 is:
Ksp = [Pb2+][Cl-]^2
Ksp = (1.73 × 10^-4 mol/L) × (8.67 × 10^-5 mol/L)^2
Ksp ≈ 1.212 × 10^-12 (rounded to three significant figures)
So, the solubility product constant (Ksp) for PbCl2 is approximately 1.212 × 10^-12.