JAMB 2024 Chemistry · Question 9

Aqueous solution of sodium hydroxide can be used to test for the presence of : I. Ca2+^{2+} ,   II. Zn2+^{2+} ,  III. Cu2+^{2+}

Options

  1. A.I only
  2. B.I and III
  3. C.I, II and III
  4. D.I and II only

Answer: C. I, II and III

Explanation written with AI and checked against the answer key

Analysis of the Question

The question asks us to identify which of the given metal ions (Ca2+Ca^{2+}, Zn2+Zn^{2+}, and Cu2+Cu^{2+}) can be identified using an aqueous solution of sodium hydroxide (NaOHNaOH). In qualitative analysis, NaOHNaOH is used as a reagent to form characteristic precipitates with various metal cations.

Explaining the Concept

When aqueous NaOHNaOH is added to solutions containing these metal ions, they react to form metal hydroxides. We look for distinct physical changes, such as the color of the precipitate or its solubility in excess NaOHNaOH.

  1. Ca2+Ca^{2+} (Calcium ion): When NaOHNaOH is added, it forms a white precipitate of calcium hydroxide: Ca(aq)2++2OH(aq)−→Ca(OH)2(s)Ca^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Ca(OH)_{2(s)} (White precipitate)
  2. Zn2+Zn^{2+} (Zinc ion): When NaOHNaOH is added, it forms a white precipitate of zinc hydroxide. Crucially, this precipitate is amphoteric and dissolves in excess NaOHNaOH to form a colorless solution of sodium tetrahydroxozincate(II): Zn(aq)2++2OH(aq)−→Zn(OH)2(s)Zn^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Zn(OH)_{2(s)} (White precipitate) Zn(OH)2(s)+2OH(aq)−→[Zn(OH)4](aq)2−Zn(OH)_{2(s)} + 2OH^-_{(aq)} \rightarrow [Zn(OH)_4]^{2-}_{(aq)} (Colorless solution)
  3. Cu2+Cu^{2+} (Copper(II) ion): When NaOHNaOH is added, it forms a characteristic pale blue precipitate of copper(II) hydroxide: Cu(aq)2++2OH(aq)−→Cu(OH)2(s)Cu^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Cu(OH)_{2(s)} (Pale blue precipitate)
Step-by-Step Solution
  • Step 1: Test for Ca2+Ca^{2+}: NaOHNaOH produces a white precipitate. This confirms the presence of Ca2+Ca^{2+}.
  • Step 2: Test for Zn2+Zn^{2+}: NaOHNaOH produces a white precipitate that dissolves in excess. This confirms the presence of Zn2+Zn^{2+}.
  • Step 3: Test for Cu2+Cu^{2+}: NaOHNaOH produces a pale blue precipitate. This confirms the presence of Cu2+Cu^{2+}.

Since all three ions produce distinct, observable reactions with aqueous sodium hydroxide, NaOHNaOH is a valid reagent for testing the presence of all three.

Conclusion

Therefore, the correct option is C.

Exam Tip

When dealing with qualitative analysis questions, always remember the "Amphoteric Trio": Zinc (Zn2+Zn^{2+}), Lead (Pb2+Pb^{2+}), and Aluminium (Al3+Al^{3+}). These three form white precipitates with NaOHNaOH that dissolve in excess. Knowing this will help you quickly identify them in both JAMB and WASSCE practical-based theory questions!

AI can make mistakes. Report anything that looks wrong.

Past questions © the examination body (JAMB/WAEC/NECO). DeeepSolve is not affiliated with them. Rights holders: see /copyright.