How do you write the Ksp expression for lead chromate (#PbCrO_4#) and calculate its solubility in mol/L? Ksp= #2.3 *10^-13#
You can do it like this:
In a saturated solution of lead chromate the solid is in equilibrium with its ions:
This is temperature dependent and the temperature should have been specified in the question.
From the equilibrium you can see that
Since lead chromate is 1 molar with respect to lead(II) ions, then this is also equal to the solubility at that particular temperature.