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?

Question

3679 views around the world

You can reuse this answer
Creative Commons License

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?