Deciding Whether a Reaction Will Precipitate or Dissolve: Competing Equilibria
A 100 mL sample of industrial wastewater contains Mg²⁺ at 0.010 mol/L and Ca²⁺ at 0.0050 mol/L. The plant adds NaOH to raise the pH to precipitate the metals as hydroxides. Ksp(Mg(OH)₂) = 5.6 × 10⁻¹²; Ksp(Ca(OH)₂) = 5.5 × 10⁻⁶. The plant aims to reduce [Mg²⁺] to below 1.0 × 10⁻⁵ mol/L before discharge. A technician proposes adding solid Na₂CO₃ instead, since Ksp(CaCO₃) = 3.3 × 10⁻⁹.
1. Calculate the pH at which Mg²⁺ begins to precipitate and the pH needed to reach the target [Mg²⁺], and state whether Ca(OH)₂ will precipitate at that pH.
Read the worked answer
Onset of Mg(OH)₂ precipitation: Ksp = [Mg²⁺][OH⁻]² → [OH⁻] = √(5.6 × 10⁻¹²/0.010) = √(5.6 × 10⁻¹⁰) = 2.37 × 10⁻⁵ mol/L → pOH = 4.63 → pH = 9.37.
Target [Mg²⁺] = 1.0 × 10⁻⁵: [OH⁻] = √(5.6 × 10⁻¹²/1.0 × 10⁻⁵) = √(5.6 × 10⁻⁷) = 7.48 × 10⁻⁴ mol/L → pOH = 3.13 → pH = 10.87.
Ca(OH)₂ check at pH 10.87: [OH⁻] = 7.48 × 10⁻⁴. Ion product = [Ca²⁺][OH⁻]² = 0.0050 × (7.48 × 10⁻⁴)² = 0.0050 × 5.6 × 10⁻⁷ = 2.8 × 10⁻⁹, far below Ksp = 5.5 × 10⁻⁶. Ca(OH)₂ does not precipitate - the solution is undersaturated by a factor of ~2000. So Mg can be removed selectively at pH ≈ 10.9 without losing calcium.
2. Evaluate the Na₂CO₃ proposal: will it remove Mg²⁺ effectively? Support with a calculation.
Read the worked answer
Na₂CO₃ precipitates carbonates, and MgCO₃ is moderately soluble (Ksp ≈ 6.8 × 10⁻⁶), whereas CaCO₃ is very insoluble (3.3 × 10⁻⁹).
For Ca²⁺: precipitation begins when [CO₃²⁻] = Ksp/[Ca²⁺] = 3.3 × 10⁻⁹/0.0050 = 6.6 × 10⁻⁷ mol/L - very easy to reach.
For Mg²⁺: to reach the target [Mg²⁺] = 1.0 × 10⁻⁵, required [CO₃²⁻] = 6.8 × 10⁻⁶/1.0 × 10⁻⁵ = 0.68 mol/L - a very high carbonate concentration, far more than is practical and far more than needed for Ca.
Conclusion: Na₂CO₃ is selective for Ca²⁺, not Mg²⁺. It would remove calcium first and leave magnesium largely in solution, the opposite of the plant's goal. The NaOH route is correct; Na₂CO₃ could be used afterwards to polish residual Ca²⁺.
3. The wastewater also contains 0.0010 mol/L NH₄⁺, which consumes OH⁻. Explain qualitatively how this affects the required NaOH dose and whether the target is still achievable.
Read the worked answer
NH₄⁺ is a weak acid (Ka ≈ 5.6 × 10⁻¹⁰) and will react with added OH⁻: NH₄⁺ + OH⁻ → NH₃ + H₂O. This buffers the solution and consumes base, so more NaOH is needed to reach the same pH.
Quantitatively, 0.0010 mol/L NH₄⁺ in 100 mL = 1.0 × 10⁻⁴ mol, which will consume roughly that much OH⁻ before the pH can rise past the NH₄⁺/NH₃ buffer region (pKa ≈ 9.25) - close to the Mg precipitation onset (pH 9.37), so the buffer competes directly.
However, once NH₄⁺ is converted to NH₃ (which is volatile and can be stripped or oxidised), continued NaOH addition still raises the pH to 10.9 and the target is achievable - it simply requires extra base to overcome the buffer. The plant should either account for the NH₄⁺ in the dose calculation or remove ammonia first (air stripping) to save reagent.