The solubility equilibrium constant for the dissolution of copper (I) chloride into copper ions and chlorine ion is Ksp = [Cu+][Cl-].

Question

The solubility equilibrium constant for the dissolution of copper (I) chloride into copper ions and chlorine ion is K_sp = [Cu⁺][Cl^-].

If the solubility constant is K_sp = 1.7 × 10^-7, and the concentration of chlorine (Cl^-) is 0.2 mol/L greater that the concentration of copper (Cu⁺), what is the concentration of copper ions in solution? Round your answer to 2 significant figures.

The concentration of copper ions in solution is

______ mol/L.

Answer 1

CuCl(s) <==> Cu⁺(aq) + Cl^-(aq)

Ksp = [Cu⁺][Cl^-] = 1.7x10^-7

Let [Cu⁺] = x then [Cl^-] = [x + 0.2]

1.7x10^-7 = (x)(0.2 + x)

1.7x10^-7 = 0.2x + x = 1.2x

x = 1.4x10-⁷ M = [Cu⁺]