Question

The solubility of lead chromate (PbCrO4) is 4.5 x 10-5 g/L. Calculate the solubility product of this compound.

a) 4.5×10 −5

b) 2.25×10 −5

c) 9.0×10 −10

d) 2.0×10 −10

Answer 1

To find the Ksp of lead chromate, convert the given solubility from grams per liter to molarity, then square this molarity since both ions have a 1:1 stoichiometry. The calculated solubility product is 1.96 x 10^-14, which doesn't match the provided options, indicating a possible need for further examination.

Step-by-step explanation:

To calculate the solubility product (Ksp) of lead chromate (PbCrO4), we must consider the dissolution stoichiometry and the solubility given in grams per liter. For PbCrO4, which dissociates into Pb2+ and CrO42- in a 1:1 molar ratio, the molar solubility can be determined first. Given that the solubility of PbCrO4 is 4.5 x 10-5 g/L, we convert this to moles per liter (molarity) using the molecular weight of PbCrO4 (323.2 g/mol).

First, convert grams per liter to moles per liter:

Molar solubility = (4.5 x 10-5 g/L) / (323.2 g/mol) = 1.4 x 10-7 M

Now, we use this to find the concentrations of the ions in solution:

[Pb2+] = [CrO42-] = 1.4 x 10-7 M

Finally, we calculate the solubility product (Ksp):

Ksp = [Pb2+][CrO42-] = (1.4 x 10-7 M)2 = 1.96 x 10-14

However, since this product is not among the provided options, we should check the stoichiometry or see if an approximation is needed.