# Comparison of the solubility of lead(II) iodide in water and in lead(II) nitrate solution [closed]

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?

• You have a chemistry exam. This doesn't look like math. – Karl Aug 22 '20 at 20:05
• @Karl whay is the problem? – user782709 Aug 22 '20 at 20:08
• 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. – Karl Aug 22 '20 at 20:25
• Did you ever heard of common ion effect? – Mathew Mahindaratne Aug 22 '20 at 20:32
• 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. – 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.

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