CuI(s) rightleftharpoonsCu^(+) + I^(-)CuI(s)⇌Cu++I−
This is an equilibrium reaction, and at 2525 ""^@C∘C we know that:
K_"sp"=1.3xx10^-12=[Cu^+][I^-]Ksp=1.3×10−12=[Cu+][I−].
As you probably know, [CuI(s)][CuI(s)] does not appear in the equilibrium expression, in that a SOLID cannot express a concentration.
If we write S=[CuI(aq)]=[Cu^+]=[I^-]S=[CuI(aq)]=[Cu+]=[I−], then........
K_"sp"=1.3xx10^-12=[Cu^+][I^-]=SxxS=S^2Ksp=1.3×10−12=[Cu+][I−]=S×S=S2.
So S=sqrt(1.3xx10^-12)=1.14xx10^-6*mol*L^-1S=√1.3×10−12=1.14×10−6⋅mol⋅L−1, with respect to "cuprous iodide"cuprous iodide.
And thus "mass of CuI"=1.14xx10^-6*cancel(mol*L^-1)xx1.2*cancelLxx190.45*g*cancel(mol^-1)=0.26*mg.
In a solution of "sodium iodide", would "cuprous iodide" be MORE or LESS soluble than in this saturated solution? Why?
