The molar solubility of Zn(OH) in 0.1 M NaOH solution is x × 10 M. The value of x is _____(Nearest i — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Write the dissolution equilibrium and Ksp expression** Zn(OH)₂(s) ⇌ Zn²⁺(aq) + 2OH⁻(aq) Ksp = [Zn²⁺][OH⁻]² = 2 × 10⁻²⁰ **Step 2: Identify the molar solubility relationship** Let the molar solubility of Zn(OH)₂ be s M. Then: [Zn²⁺] = s and [OH⁻] from Zn(OH)₂ = 2s **Step 3: Account for the common ion effect** Since NaOH is present at 0.1 M, the total [OH⁻] is approximately: [OH⁻] ≈ 0.1 + 2s ≈ 0.1 M (since s is very small due to low Ksp) **Step 4: Substitute into Ksp expression** Ksp = [Zn²⁺][OH⁻]² 2 × 10⁻²⁰ = s × (0.1)² 2 × 10⁻²⁰ = s × 10⁻² **Step 5: Solve for molar solubility s** s = (2 × 10⁻²⁰)/(10⁻²) s = 2 × 10⁻¹⁸ M **Step 6: Identify the value of x** Since s = x × 10⁻¹⁸ M, we have x = 2 Therefore, the answer is **2.00**.