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)#....