# Question #69c9d

Apr 12, 2015

To determine the solubility of zinc cyanide you need the value of its solubility product constant, ${K}_{s p}$, which is listed at being $8.0 \cdot {10}^{- 12}$.

So, zinc cyanide will not dissociate completely in aqueous solution; however, you can use the ICE table for the equilibrium reaction that takes place

$\text{ } Z n \left(C N\right) 2 \left(s\right) r i g h t \le f t h a r p \infty n s Z {n}_{\left(a q\right)}^{2 +} + \textcolor{red}{2} C {N}_{\left(a q\right)}^{-}$
I............$-$........................0...............0.
C..........$-$.......................(+x)..........(+$\textcolor{red}{2}$x)
E...........$-$........................x...............2x

Notice that, at equilibrium, you get 1 mole of $Z {n}^{2 +}$ cations and $\textcolor{red}{2}$ moles of ${\left(C N\right)}^{-}$ anions. The solubility product constant, ${K}_{s p}$ for this reaction will be

${K}_{s p} = \left[Z {n}^{2 +}\right] \cdot {\left[C {N}^{-}\right]}^{\textcolor{red}{2}}$

${K}_{s p} = x \cdot {\left(2 x\right)}^{2} = 4 {x}^{3} = 8.0 \cdot {10}^{- 12}$

Solve for $x$ and you'll get

$x = \textcolor{g r e e n}{1.26 \cdot {10}^{- 4} \text{mol/L}}$ $\to$ this is zinc cyanide's molar solubility.

Now, to get its solubility in grams per liter, all you have to do is multiply its molar solubility by its molar mass

$1.26 \cdot {10}^{- 4} \cancel{\text{mol")/"L" * "117.44 g"/(1cancel("mole")) = color(green)(1.48 * 10^(-2)"g/L}}$