-1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Yep. I think it's insoluble, so the concentration would be zero? $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Dec 9 '15 at 2:23
  • 1
    $\begingroup$ No, it's not completely insoluble, it's just after a certain point, it cannot be saturated any further. Try searching solubility problems regarding Ksp to better understand the problem, try this video: link. $\endgroup$ – Yulmart Dec 9 '15 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.