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

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#

Serenity Allen · Accepted Answer

You can do it like this:

In a lead chromate saturated solution, the solid and its ions are in balance: #sf(PbCrO_(4(s))rightleftharpoonsPb_((aq))^(2+)+CrO_(4(aq))^(2-))# For what: #sf(K_((sp))=[Pb_((aq))^(2+)][CrO_(4(aq))^(2-)]=2.3xx10^(-13)" ""mol"^2."l"^(-2))# Since this depends on temperature, the question ought to have asked what the temperature was. 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")# Lead chromate's solubility at that specific temperature is also equal to its 1 molar relationship with lead(II) ions.

Jose Ayala · Answer

undefined $$ \text{K}_{\text{sp}} = [\text{Pb}^{2+}][\text{CrO}_4^{2-}] $$

$$ \text{K}_{\text{sp}} = (x)(x) = x^2 $$

$$ x = \sqrt{\text{K}_{\text{sp}}} = \sqrt{2.3 \times 10^{-13}} $$

$$ \text{Solubility} = [\text{Pb}^{2+}] = x = \sqrt{2.3 \times 10^{-13}} $$