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

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Answer 1 · 2016-07-03T17:11:23

You can do it like this:

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.

How do you write the Ksp expression for lead chromate (PbCrO_4PbCrO_4) and calculate its solubility in mol/L? Ksp= 2.3 *10^-132.3 *10^-13