Lead chloride dissolves in water according to
PbCl2(s)↔Pb2++2Cl−(aq)
The solubility in pure water has been measured to be 4.44 g⋅L−1. Calculate the solubility product of lead chloride in pure water. Your answer should be in units of molarity raised to the appropriate exponent.
To calculate the solubility product of lead chloride (PbCl2) in pure water, we need to use the given solubility (4.44 g⋅L−1) and the balanced equation for the dissolution of lead chloride:
PbCl2(s)↔Pb2+(aq) + 2Cl−(aq)
The solubility product constant (Ksp) expression for this reaction is:
Ksp = [Pb2+][Cl−]^2
To determine the concentrations of the ions, we need to convert the solubility of PbCl2 to molar concentration. The molar mass of PbCl2 is 278.1 g/mol, so:
1 mol PbCl2 = 278.1 g PbCl2
Now, we can find the molar concentration of PbCl2 as follows:
4.44 g/L ÷ 278.1 g/mol = 0.0159 mol/L
Since 1 mol of PbCl2 yields 1 mol of Pb2+ and 2 mol of Cl−, the concentration of Pb2+ in the solution is also 0.0159 mol/L. The concentration of Cl− is twice as much, so it is 2 × 0.0159 mol/L = 0.0318 mol/L.
Now we have the concentrations of Pb2+ and Cl−, we can substitute these values into the solubility product constant expression:
Ksp = (0.0159 mol/L)(0.0318 mol/L)^2
Calculating this expression gives:
Ksp = 1.61 x 10^−5 mol^3/L^3
Therefore, the solubility product of lead chloride in pure water is 1.61 x 10^−5 mol^3/L^3.
The solubility product (Ksp) is a measure of the equilibrium concentration of the ions in a saturated solution. It can be calculated by multiplying the concentrations of the ions raised to the power of their stoichiometric coefficients in the balanced equation.
In the given equation, the stoichiometric coefficients for Pb2+ and Cl- are 1 and 2 respectively. Therefore, the Ksp for lead chloride (PbCl2) can be calculated as follows:
Ksp = [Pb2+][Cl-]^2
Since the solubility of lead chloride in pure water is given as 4.44 g/L, we need to convert this to molarity. The molar mass of lead chloride (PbCl2) is calculated as follows:
Molar mass of PbCl2 = (atomic mass of Pb) + 2 * (atomic mass of Cl) = 207.2 g/mol + 2 * 35.45 g/mol = 278.1 g/mol
Next, we can calculate the molarity (M) of lead chloride using the given solubility:
[Lead chloride] = (4.44 g/L) / (278.1 g/mol) = 0.016 M
Now, substituting the values into the equation for Ksp:
Ksp = (0.016 M) * (0.016 M)^2 = 0.000042 M^3
Therefore, the solubility product (Ksp) of lead chloride in pure water is 0.000042 M^3.