When sodium hydroxide solution is added to a solution of zinc salt, the white precipitate formed re-dissolve in excess sodium hydroxide because

Step 1: Analyze the question When NaOH is added to a zinc salt solution, a white precipitate forms, and it dissolves when more NaOH is added. We need the reason for this.

Step 2: The concept Some metal hydroxides are amphoteric. This means they react with both acids and strong bases. Zn(OH)₂, Al(OH)₃ and Pb(OH)₂ are the common examples in the syllabus.

Step 3: Solution

1. A small amount of NaOH gives the white precipitate of zinc hydroxide: Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)

2. In excess NaOH, the zinc hydroxide acts as an acid and reacts with the base. This forms a soluble sodium zincate (or tetrahydroxozincate) complex: Zn(OH)₂(s) + 2NaOH(aq) → Na₂ZnO₂(aq) + 2H₂O(l) or Zn(OH)₂(s) + 2OH⁻(aq) → [Zn(OH)₄]²⁻(aq)

The precipitate disappears and gives a colorless solution.

Step 4: Check the options

• A. A strong alkali alone does not explain it, because NaOH does not dissolve Mg(OH)₂ or Fe(OH)₂.
• B and C. Reactivity of the metals has nothing to do with the dissolving of the precipitate.
• D. Zinc hydroxide is amphoteric. It reacts with the excess base, so it dissolves. This is correct.
• E. Instability would mean it decomposes, not dissolves.

Exam tip: Remember "Zn, Al, Pb", the metals whose hydroxides dissolve in excess NaOH. Zn(OH)₂ also dissolves in excess aqueous ammonia, but Al(OH)₃ and Pb(OH)₂ do not. This difference is a common way to tell them apart in WASSCE and NECO qualitative analysis.