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.