When a solution of Cu(II) chloride, CuCl₂ , is added to a solution of potassium sulfide, K₂S, a black precipitate of copper sulfide, CuS, forms. Which of the following is the total ionic equation for this reaction?

Question

Quizplus · Accepted Answer

The ionic equation represents the dissociation of all soluble ionic species into their constituent ions, leaving only the insoluble precipitate(s) in their solid form. In the reaction given, Cu(II) chloride dissociates into Cu(II) and Cl- ions, while potassium sulfide dissociates into K+ and S2- ions. The Cu(II) ions and S2- ions combine to form the insoluble precipitate of copper sulfide, leaving K+ ions and Cl- ions in solution.

The correct ionic equation is:

Cu2+ (aq) + S2- (aq) + 2K+ (aq) + 2Cl- (aq) → CuS (s) + 2K+ (aq) + 2Cl- (aq)

Simplifying the equation by canceling out the spectator ions (K+ and Cl- ions) present on both sides of the equation, we get:

Cu2+ (aq) + S2- (aq) → CuS (s)

Thus, the total ionic equation is:

CuCl2 (aq) + K2S (aq) → CuS (s) + 2KCl (aq)

Option C correctly represents this equation by breaking down all the ionic species into their constituent ions and showing them in their ionic form. Option A is incorrect because it shows only the species that are directly involved in the reaction and does not include the spectator ions. Option B is incorrect because it shows unnecessary additional species and is overly complicated. Option D is incorrect because it does not show any ionic species and only shows the reactant and product as molecular species.

When a Solution of Cu(II) Chloride, CuCl2 , Is Added

When a Solution of Cu(II) Chloride, CuCl₂ , Is Added