Question

Part a. 1. NH₃ is a weak base; NaOH is a strong base. Predict what would appear in the solution if NaOH had been added to the CuSO₄ solution instead of the NH₃. (Hint: see Appendix E.)

a) Copper precipitate
b) Ammonium sulfate
c) Sodium sulfate
d) Hydrogen gas

Answer 1

When NaOH is added to CuSO4, sodium sulfate remains dissolved, and a precipitate of copper (II) hydroxide will form because NaOH is a strong base that completely dissociates, leading to a reaction with copper ions.

Step-by-step explanation:

If NaOH had been added to the CuSO₄ solution instead of NH₃ (ammonia), we expect the strong base NaOH to react with CuSO₄ to form a precipitate of copper (II) hydroxide, which is insoluble in water. Unlike NH₃, which forms a complex with copper ions in solution, NaOH will increase the hydroxide ion concentration significantly due to its complete ionization, as indicated by the reaction:

NaOH(s) → Na⁺ (aq) + OH⁻ (aq)

This reaction is almost complete because NaOH is a strong base. In the presence of copper sulfate, the hydroxide ions (OH⁻) will react with copper (II) ions (Cu²⁺) to form copper (II) hydroxide (Cu(OH)₂), which is a blue precipitate. The resulting product in the solution would be sodium sulfate (Na₂SO₄) as a soluble salt. Therefore, the correct answer to the student's question is c) Sodium sulfate.