Given that $K_"sp"=1.3xx10^(-12)$ for $"cuprous iodide"$ , $CuI$ , what mass of this salt would dissolve in a $1.2*L$ volume of water?

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Answer 1 · 2017-03-23T15:22:27

We assess the equilibrium:

$CuI(s) rightleftharpoonsCu^(+) + I^(-)$

$CuI(s) rightleftharpoonsCu^(+) + I^(-)$

This is an equilibrium reaction, and at $25$ $""^@C$ we know that:

$K_"sp"=1.3xx10^-12=[Cu^+][I^-]$ .

As you probably know, $[CuI(s)]$ does not appear in the equilibrium expression, in that a SOLID cannot express a concentration.

If we write $S=[CuI(aq)]=[Cu^+]=[I^-]$ , then........

$K_"sp"=1.3xx10^-12=[Cu^+][I^-]=SxxS=S^2$ .

So $S=sqrt(1.3xx10^-12)=1.14xx10^-6*mol*L^-1$ , with respect to $"cuprous iodide"$ .

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?

Given that K_"sp"=1.3xx10^(-12)K_"sp"=1.3xx10^(-12) for "cuprous iodide""cuprous iodide", CuICuI, what mass of this salt would dissolve in a 1.2*L1.2*L volume of water?