Question

Use equilibrium equations to explain why Cu(Cl)2 is more soluble in a solution of sodium cyanide than in pure water, given that it forms the complex ion Cu(CN)42-.

Answer 1

CuCl₂ is sparingly soluble in water give the solubility equilibrium,

CuCl₂ <=> Cu²⁺ + 2Cl^- --------------------(1)

In cyanide solution, Cu complexes with the cyanide ion and forms Cu(CN)₄^2-

   Cu²⁺ + 4CN^- -> Cu(CN)₄^2-

The dissolved CuCl₂ forms Cu²⁺ which forms  Cu(CN)₄^2-, which affects the equilibrium shown in equation (1)

According to Le Chatelier's principle, if concentration of product in an equilibrium is reduced, then the forward reaction speeds up.

Here The forward reaction is the dissolution process. So as Cu²⁺ gets taken to form the complex, the equilibrium makes the forward reaction more fast, hence dissolution of Cu²⁺ happens and CuCl₂ gets more dissolved in NaCN solution.