How many equivalents of iodide anion compared to lead cations in #PbI_2#?

1 Answer
anor277
Aug 21, 2015

Your answer is in the question; that is, if you have transcribed the question correctly.

Explanation:

#PbI_2(s) rightleftharpoons Pb^(2+) + 2I^-#

You have been quoted the concentration of plumbic ions, #Pb^(2+)#. It is easy to see why this concentration is equal to the required solubility. As a followup to this question, what do you think the examiner means by "saturated solution"?

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