Question

Calculate the equilibrium constant for the following reaction. Zn(OH)2(s) + 2CN-(aq) <--->Zn(CN)2(s) + 2 OH-(aq) Does...

Calculate the equilibrium constant for the following reaction.
Zn(OH)2(s) + 2CN-(aq) <--->Zn(CN)2(s) + 2 OH-(aq)

Does the equilibrium lie predominately to the left or to the right? Can zinc hydroxide be transformed into zine cyanide by adding a soluble salt of the cyanide ion?Calculate the equilibrium constant for the following reaction.

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Answer #1

Ksp(Zn(OH)2) = 4.5x10-17
Ksp(Zn(CN)2) = 8.0x10-12
Therefore the equilibrium constant, Keq , for the given reaction as written is
Keq = Ksp(Zn(OH)2) / Ksp(Zn(CN)2)
Keq = 4.5x10-17 / 8.0x10-12 = 5.6x10-6

K = [OH-]² / [CN-]²
Since K<1, then the equilibrium lies predominantly to the left.

According to Le Chatelier's principle, adding soluble CN- to the reaction mixture will drive the reaction to the right (no matter what K is, as long as equilibrium is established). So, it is possible to transform zinc hydroxide into zinc cyanide

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