Solubility of $\ce{PbI2}$: Is $\ce{PbI2}$ more soluble in $\ce{H2O}$ or in an aqueous solution of Lead(II) nitrate?
$K_\mathrm{sp} (\ce{PbI_{2}})=8.7\times 10^{-9}$

I'm not very good in Chemistry, I study mathematics and now I have this exam (in chemistry), but I have no idea how to proceed in this exercise. Can someone explain the method to solve it?

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    $\begingroup$ You have a chemistry exam. This doesn't look like math. $\endgroup$ – Karl Aug 22 '20 at 20:05
  • $\begingroup$ @Karl whay is the problem? $\endgroup$ – user782709 Aug 22 '20 at 20:08
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    $\begingroup$ If you are supposed to be able to answer this, then you must know a bit more. You're welcome to ask here if you encounter a problem when revising your lecture notes, but you must show your own effort. $\endgroup$ – Karl Aug 22 '20 at 20:25
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    $\begingroup$ Did you ever heard of common ion effect? $\endgroup$ – Mathew Mahindaratne Aug 22 '20 at 20:32
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    $\begingroup$ HINT - Start at the beginning. Write out the chemical equations for lead iodide and lead nitrate dissolving in water. Then think about the common ion effect which is a specific case of Le Chatelier's principle. $\endgroup$ – MaxW Aug 22 '20 at 20:53

In water the following equilibrium would establish:

$$\ce{PbI2 (s) <=> Pb^2+ (aq) + 2I- (aq) \tag1}$$

$$\therefore \ K_\mathrm{sp} = [\ce{Pb^2+}][\ce{I-}]^2 = 8.7 \times 10^{−9} \tag2$$

Now, using equations $(1)$ and $(2)$, you can calculate $[\ce{Pb^2+}]$ and $[\ce{I-}]$ in water.

In $\ce{Pb(NO3)2}$ solution, following ions are present:

$$\ce{Pb(NO3)2 (aq) -> Pb^2+ (aq) + 2NO3- (aq) \tag3}$$

Thus, $\ce{Pb^2+}$ is present abundantly in the solution as the common ion (suppose $[\ce{Pb^2+}]$ is $\pu{0.1 M}$). Still, the equilibrium representing the equation $(1)$ should establish regardless of the presence of common ion. However, the equilibrium concentrations of $[\ce{Pb^2+}]$ and $[\ce{I-}]$ are very different than that of with pure water you calculated before. Yet, $K_\mathrm{sp}$ is still the same. So, you can again calculate $[\ce{I-}]$ using the equation $(2)$ and known $K_\mathrm{sp}$. Here, $[\ce{Pb^2+}] \approx \pu{0.1 M}$. You should think about the reson why we are using $[\ce{Pb^2+}] = \pu{0.1 M}$ here.

Once you find $[\ce{I-}]$ in water and in $\ce{Pb(NO3)2}$ solution, respectively, you would be able to see which solution makes $\ce{PbI2}$ more soluble.

  • $\begingroup$ If i didn't do it wrong, I think that PbI_{2} is more soluble in water. Am I right? $\endgroup$ – user782709 Aug 23 '20 at 1:26
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    $\begingroup$ Excellent! Keep up with your work. Chemistry is not that bad. :-) $\endgroup$ – Mathew Mahindaratne Aug 23 '20 at 4:04

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