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Question

Copper(I) ions in aqueous solution react with NH₃(aq) according to

Cu⁺(aq) + 2NH₃(aq) ⟶ Cu(NH₃)₂⁺(aq)? f = 6.3×10¹⁰

Calculate the solubility (in g·L⁻¹) of CuBr(s) (?sp=6.3×10⁻⁹) in 0.65M NH₃(aq).

Answer “Copper(i) ions in aqueous solution react with NH₃(aq) according to”.

Effect of Complex Ion Formation on Solubility:

An electron donor ligand can form a complex ion when it is dissolved with one of the ions from the dissolution a sparinglysoluble salt. This increases the solubility of the sparinglysoluble salt. Consuming the transition metal ion causes the dissolution the sparingly-soluble salt.

Answer and Explanation:

Below are the equilibria of the given reaction and the overall chemical equation.

Cu⁺(aq) + 2NH₃(aq) ⇌ Cu(NH₃)₂⁺(aq)Kf = 6.3 x 10¹⁰

CuBr(s) ⇌ Cu⁺(aq) + Br⁻(aq) Ksp = 6.4×10⁻⁹

CuBr(s) + 2NH₃(aq) ⇌ Cu(NH₃)₂⁺(aq) + Br⁻(aq) Koverall = 403.2

The RICE table can be used to determine the equilibrium concentrations of bromide ions.

We get the Equilibrium Constant expression by establishing it.

We can solve the x value by taking the square root from both sides.

Thus, we can determine the mass solubility of CuBr based on the amount of Bromide ions present.

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