# 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?

Mar 23, 2017

We assess the equilibrium:

$C u I \left(s\right) r i g h t \le f t h a r p \infty n s C {u}^{+} + {I}^{-}$

#### Explanation:

$C u I \left(s\right) r i g h t \le f t h a r p \infty n s C {u}^{+} + {I}^{-}$

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

${K}_{\text{sp}} = 1.3 \times {10}^{-} 12 = \left[C {u}^{+}\right] \left[{I}^{-}\right]$.

As you probably know, $\left[C u I \left(s\right)\right]$ does not appear in the equilibrium expression, in that a SOLID cannot express a concentration.

If we write $S = \left[C u I \left(a q\right)\right] = \left[C {u}^{+}\right] = \left[{I}^{-}\right]$, then........

${K}_{\text{sp}} = 1.3 \times {10}^{-} 12 = \left[C {u}^{+}\right] \left[{I}^{-}\right] = S \times S = {S}^{2}$.

So $S = \sqrt{1.3 \times {10}^{-} 12} = 1.14 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$, with respect to $\text{cuprous iodide}$.

And thus $\text{mass of CuI} = 1.14 \times {10}^{-} 6 \cdot \cancel{m o l \cdot {L}^{-} 1} \times 1.2 \cdot \cancel{L} \times 190.45 \cdot g \cdot \cancel{m o {l}^{-} 1} = 0.26 \cdot m g$.

In a solution of $\text{sodium iodide}$, would $\text{cuprous iodide}$ be MORE or LESS soluble than in this saturated solution? Why?