The solubility product constant (Ksp) for the dissolution of PbSO4 as represented by the chemical equation is 1.7x10-8.
PbSO4 (s)= Pb2+ (aq) + SO42- (aq)
Calculate the mass (g) of PbSO4 that dissolves in 1100 mL of water.
To calculate the mass of PbSO4 that dissolves in 1100 mL of water, we need to use the solubility product constant (Ksp) and the formula weight of PbSO4.
1. First, convert the volume of water to liters:
1100 mL ÷ 1000 mL/L = 1.1 L
2. Write the solubility equilibrium expression:
Ksp = [Pb2+][SO42-]
3. Since the dissociation of PbSO4 produces 1 mole of Pb2+ and 1 mole of SO42-, the expression becomes:
Ksp = [Pb2+][SO42-]
4. Assume x is the amount (in moles) of PbSO4 that dissolves.
5. Since 1 mole of PbSO4 dissociates to 1 mole of Pb2+, the concentration of Pb2+ in the solution will also be x moles/L.
6. The concentration of SO42- in the solution will also be x moles/L.
7. Substitute the concentrations into the solubility product expression:
Ksp = x * x
Ksp = x^2
8. Solve for x by taking the square root of Ksp:
x = √(Ksp)
x = √(1.7x10^-8)
x = 1.30x10^-4 moles/L
9. Calculate the mass of PbSO4 using the formula weight of PbSO4.
Formula weight of PbSO4 = 207.2 g/mol
Mass of PbSO4 = x moles/L * 207.2 g/mol * 1.1 L
Mass of PbSO4 = (1.30x10^-4 moles/L) * (207.2 g/mol) * 1.1 L
Mass of PbSO4 = 30.1 mg
Therefore, the mass of PbSO4 that dissolves in 1100 mL of water is 30.1 mg.
To calculate the mass of PbSO4 that dissolves in 1100 mL of water, we need to use the solubility product constant and the volume of water.
1. Start by converting the volume of water to liters.
1100 mL = 1100/1000 = 1.1 L
2. Write down the chemical equation representing the dissolution of PbSO4.
PbSO4 (s) ⇌ Pb2+ (aq) + SO42- (aq)
3. Use the solubility product constant (Ksp) to set up an expression for the equilibrium solubility of PbSO4.
Ksp = [Pb2+][SO42-]
4. Since the solubility of PbSO4 will be equal to the concentration of Pb2+ and SO42- ions, we can substitute [Pb2+] = x and [SO42-]= x into the Ksp expression.
Ksp = x * x = x^2
5. Substitute the value of Ksp into the equation.
1.7x10^-8 = (x^2)
6. Solve for x by taking the square root of both sides.
x = sqrt(1.7x10^-8)
7. Calculate the solubility of PbSO4 in moles per liter (mol/L).
solubility = x = sqrt(1.7x10^-8) mol/L
8. Calculate the amount of PbSO4 in moles that dissolves in 1.1 L of water.
moles of PbSO4 = solubility * volume of water in liters
= sqrt(1.7x10^-8) * 1.1 mol
9. Convert moles to grams using the molar mass of PbSO4.
molar mass of PbSO4 = atomic mass of Pb + atomic mass of S + (4 * atomic mass of O)
= (207.2 g/mol) + (32.1 g/mol) + (4 * 16.0 g/mol)
= 303.3 g/mol
mass of PbSO4 = moles of PbSO4 * molar mass of PbSO4
Therefore, plugging in the values and calculating:
mass of PbSO4 = sqrt(1.7x10^-8) * 1.1 * 303.3 g
This will give you the mass (in grams) of PbSO4 that dissolves in 1100 mL of water.
Pb^+2 = x
SO4^-2 = x
Substitute into Ksp and solve.
Then convert M into g/1100 mL.
What about this question?
Calculate the mass (g) of PbSO4 that will dissolve in 2000 mL of 0.0694 M K2SO4 solution. The equation represents the solubility process.
PbSO4 (s) Pb2+ (aq) + SO42- (aq)
Ksp = 1.7 x 10-8