Amphoteric vs. basic hydroxides in excess NaOH

Steps:

• Adding NaOH precipitates both Zn(OH)₂ (white) and Cu(OH)₂ (blue) initially.
• Excess NaOH dissolves amphoteric Zn(OH)₂ as [Zn(OH)₄]²⁻ (zincate ion), leaving it in solution.
• Cu(OH)₂ remains insoluble as it is not amphoteric.
• Filtration retains only the insoluble blue Cu(OH)₂ on the filter paper.

Why C is correct:

• Cu(OH)₂ is a basic hydroxide insoluble in excess NaOH, forming 0.1 mol of blue precipitate per the reaction Cu²⁺ + 2OH⁻ → Cu(OH)₂(s).

Why the others are wrong:

• A: Zn(OH)₂ dissolves in excess NaOH, so no white hydroxide remains.
• B: Zn(OH)₂ is white but amphoteric and dissolves, leaving no residue.
• D: Cu(OH)₂ precipitates, so solid residue forms.

Final answer: C