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.
The solubility of lead chromate is 4.5×10^-5. Calculate the solubility product of this compound
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