What is the ratio of $[Pb^2+]$ to $[I^-]$ in a saturated solution of $PbI_2$ ?

Question

Which of the following is true for a saturated solution of lead (II) iodide, a slightly soluble salt?

A. $[Pb^2+] = [I^-]$
B. $[Pb^2+] = K_(sp)$
C. $[Pb^2+] = 0.5[I^-]$
D. $[Pb^2+] = (K_(sp))^(1/2)$
E. $[Pb^2+] = 2[I^-]$

I thought it was (E) but evidently it is (C); it seems trivial but it isn't making sense!

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Answer 1 · 2017-11-09T17:52:29

Look at the equilibrium expression....

$PbI_2(s) rightleftharpoonsPb^(2+) + 2I^-$ ....where $K_"sp"=[Pb^(2+)][I^-]^2$ ...

Clearly in a solution containing ONLY $PbI_2$ as the ion source, $[Pb^(2+)]=1/2[I^-]$ ... $"option C"$ as required.... Claro?

Note that if we were required to calculate the solubility of $PbI_2$ , we would get an expression of the form....

$S_"solubility of lead(II) iodide"=""^(3)sqrt(K_"sp"/4)$ ....