Question

How do you write the Ksp expression for lead chromate (`PbCrO_4`) and calculate its solubility in mol/L? Ksp= `2.3 *10^-13`

Answer 1

You can do it like this:

Explanation:

In a saturated solution of lead chromate the solid is in equilibrium with its ions:

`sf(PbCrO_(4(s))rightleftharpoonsPb_((aq))^(2+)+CrO_(4(aq))^(2-))`

For which:

`sf(K_((sp))=[Pb_((aq))^(2+)][CrO_(4(aq))^(2-)]=2.3xx10^(-13)" ""mol"^2."l"^(-2))`

This is temperature dependent and the temperature should have been specified in the question.

From the equilibrium you can see that `sf([Pb_((aq))^(2+)]=[CrO_(4(aq))^(2-)])`

`:.` `sf([Pb_((aq))^(2+)]^(2)=2.3xx10^(-13)" ""mol"^2."l"^(-2))`

`:.` `sf([Pb_((aq))^(2+)]=sqrt(2.3xx10^(-13))=4.8xx10^(-7)" ""mol/l")`

Since lead chromate is 1 molar with respect to lead(II) ions, then this is also equal to the solubility at that particular temperature.