Concentration of Iodide Ions in water


> 
> You add 10.50 g of solid lead(II) iodide to 525 mL of water. Assuming that the lead(II) iodide does not add to the volume of the solution, what is the concentration of iodide ions in solution?
> 
> 
> 


Here’s what I have done:


$10.50\ \mathrm g\ \ce{PbI2}\cdot \frac{1\ \mathrm{mol}\ \ce{PbI2}}{461.00\ \mathrm g\ \ce{PbI2}}\cdot \frac{2\ \mathrm{mol}\ \ce{I-}}{1\ \mathrm{mol}\ \ce{PbI2}}= 0.04555\ \mathrm{mol}\ \ce{I-}$


$\frac{0.04555\ \mathrm{mol}\ \ce{I-}}{0.525\ \mathrm L\ \text{solution}} = 8.68\cdot 10^{-2}\ \text{Molar}\ \ce{I-}$


What went wrong?




I think you have to consider the solubility rules and determine how much of the lead (II) iodide will actually dissolve in that given amount of water rather than assuming all of it dissolves.


 1