The solubility of lead chromate is 4.5×10^-5. Calculate the solubility product of this compound
The solubility product (Ksp) of lead chromate (PbCrO4) can be calculated using the following equation:
Ksp = [Pb2+][CrO42-]
where [Pb2+] is the concentration of lead ions and [CrO42-] is the concentration of chromate ions.
Since the solubility of lead chromate is given as 4.5×10^-5, we can assume that the concentration of lead ions and chromate ions in equilibrium is also 4.5×10^-5. Therefore:
[Pb2+] = 4.5×10^-5 M
[CrO42-] = 4.5×10^-5 M
Substituting these values into the equation for Ksp:
Ksp = (4.5×10^-5)(4.5×10^-5) = 2.025×10^-9
Therefore, the solubility product of lead chromate is 2.025×10^-9.
To calculate the solubility product (Ksp) of lead chromate (PbCrO4), we need to determine the concentrations of the lead ions (Pb2+) and chromate ions (CrO4^2-) in a saturated solution.
The solubility of lead chromate is given as 4.5×10^-5. This value represents the concentration of each ion in moles per liter (M).
Since lead chromate dissociates into one lead ion and one chromate ion for every mole of compound, we can assume that the concentrations of Pb2+ and CrO4^2- are both equal to 4.5×10^-5 M.
Therefore, the solubility product (Ksp) of lead chromate can be calculated as follows:
Ksp = [Pb2+][CrO4^2-]
Ksp = (4.5×10^-5 M)(4.5×10^-5 M)
Ksp = 2.025×10^-9
Therefore, the solubility product of lead chromate is 2.025×10^-9.