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

Serenity Allen · Accepted Answer

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?

Jeremiah Adams · Answer

undefined To find the mass of cuprous iodide (CuI) that would dissolve in a 1.2 L volume of water, we need to use the solubility product constant (Ksp) and the volume of water.

The Ksp expression for cuprous iodide (CuI) is:
CuI ⇌ Cu⁺ + I⁻

The Ksp expression is: Ksp = [Cu⁺][I⁻]

Given Ksp = 1.3 × 10^(-12), and assuming complete dissociation,
[Cu⁺] = [I⁻] = √(Ksp) = √(1.3 × 10^(-12))

Now, we can find the molarity of Cu⁺ ions (and I⁻ ions) in the solution.

Molarity (M) = moles of solute / volume of solution (in liters)

Since the volume of water is given as 1.2 L, the moles of Cu⁺ (and I⁻) ions that dissolve can be found from the molarity and volume.

Now, we calculate the molar mass of CuI, which is approximately 190.45 g/mol.

Then, we use the molar mass and the number of moles of CuI to find the mass of CuI that would dissolve in the 1.2 L volume of water.