To determine the solubility of zinc cyanide you need the value of its *solubility product constant*, #K_(sp)#, which is listed at being #8.0 * 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

#" "Zn(CN)2(s) rightleftharpoons Zn_((aq))^(2+) + color(red)(2)CN_((aq))^(-)#

**I**............#-#........................0...............0.

**C**..........#-#.......................(+x)..........(+#color(red)(2)#x)

**E**...........#-#........................x...............2x

Notice that, at equilibrium, you get **1 mole** of #Zn^(2+)# cations and #color(red)(2)# moles of #(CN)^(-)# anions. The *solubility product constant*, #K_(sp)# for this reaction will be

#K_(sp) = [Zn^(2+)] * [CN^(-)]^(color(red)(2))#

#K_(sp) = x * (2x)^(2) = 4x^(3) = 8.0 * 10^(-12)#

Solve for #x# and you'll get

#x = color(green)(1.26 * 10^(-4)"mol/L")# #-># 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 * 10^(-4)cancel("mol")/"L" * "117.44 g"/(1cancel("mole")) = color(green)(1.48 * 10^(-2)"g/L")#