Calculate Oh and Ph for 0.035 M Na2s Solution

Introduction

When sodium sulfide dissolves in water, it dissociates into sodium ions (Na⁺) and sulfide ions (S²⁻). The sulfide ions then react with water to form bisulfide ions (HS⁻) and hydroxide ions (OH⁻), creating a basic solution.

The equilibrium constant for this reaction is approximately 1.2 × 10⁻13 at 25°C. This calculator uses this equilibrium to determine the OH⁻ concentration and resulting pH.

Calculation Method

The calculation involves solving the equilibrium equation for the reaction between sulfide ions and water. The steps are:

1. Assume the initial concentration of S²⁻ is equal to the Na₂S concentration (0.035 M)
2. Set up the equilibrium expression based on the reaction stoichiometry
3. Solve for the equilibrium concentration of OH⁻ using the equilibrium constant
4. Calculate pH from the OH⁻ concentration

For small concentrations of HS⁻, we can approximate [HS⁻] ≈ [OH⁻], simplifying the calculation.

Example Calculation

For a 0.035 M Na₂S solution:

1. Initial [S²⁻] = 0.035 M
2. Let x = equilibrium [OH⁻] = [HS⁻]
3. Equilibrium [S²⁻] = 0.035 - x
4. Substitute into equilibrium expression: 1.2 × 10⁻13 = x²/(0.035 - x)
5. For small x, 0.035 - x ≈ 0.035
6. Solve: x² = 1.2 × 10⁻13 × 0.035
7. x ≈ 0.000000000000042 M (4.2 × 10⁻¹⁴ M)
8. pOH = -log(4.2 × 10⁻¹⁴) ≈ 13.37
9. pH = 14 - 13.37 ≈ 0.63

Interpretation

The results indicate:

• OH⁻ concentration is extremely high (4.2 × 10⁻¹⁴ M)
• pH is 0.63, showing strong basicity
• The solution is highly corrosive and dangerous to handle

Always wear appropriate protective equipment when working with sodium sulfide solutions.