Estimate the solubility of lead (ll) chloride in grams per liter.
ksp= 1.7 x 10^-5
Given that
PbCl2 = Pb2+ + 2Cl-
then the Ksp = [Pb2+][Cl-]^2
the solubility of the lead(II)chloride in moles per litre is numerically equal to the concentration of [Pb2+] in moles per litre, because 1 mole of PbCl2 gives 1 mole fo Pb2+
so x moles of PbCl2 gives x moles of Pb2+ and 2x moles of Cl-
thus Ksp = (x)(2x)^2 = 1.7 x 10^-5
(I am assuming that your Ksp is appropriate for mole l-1)
so x = cube root [(1.7 x 10^-5)/4]
and find x
To estimate the solubility of lead (II) chloride (PbCl2) in grams per liter, we can use the concept of the solubility product constant (Ksp).
The Ksp value represents the equilibrium constant for the dissolution of a sparingly soluble salt. For PbCl2, the chemical equation for its dissociation is:
PbCl2 ↔ Pb2+ + 2Cl-
According to the equation, one mole of PbCl2 produces one mole of Pb2+ and two moles of Cl-. Therefore, the solubility of PbCl2 in terms of Pb2+ concentration can be represented as "s", and the concentration of Cl- ions will be 2s.
The Ksp expression for PbCl2 is then:
Ksp = [Pb2+][Cl-]^2
Substituting the concentrations, we have:
1.7 x 10^-5 = s(2s)^2
1.7 x 10^-5 = 4s^3
Solving for "s", we can take the cube root of both sides:
s^3 = (1.7 x 10^-5)/4
s^3 = 4.25 x 10^-6
Taking the cube root:
s ≈ 0.015 M
To convert this molar solubility (M) to grams per liter (g/L), we need to consider the molar mass of PbCl2, which is 278.10 g/mol. Using this information, we can convert the solubility in M to g/L as follows:
0.015 M PbCl2 × (278.10 g PbCl2 / 1 mol PbCl2) × (1 L / 1000 mL) = 0.0408 g PbCl2/L
Therefore, the estimated solubility of lead (II) chloride in grams per liter is approximately 0.0408 g/L.