Calculate The Ph of A 0.082 M Solution of Nacn.

Introduction

Sodium cyanide (NaCN) is a strong electrolyte that dissociates completely in water to form Na⁺ and CN⁻ ions. The cyanide ion (CN⁻) reacts with water to form the weak acid HCN and hydroxide ions (OH⁻).

The pH of a NaCN solution depends on the concentration of HCN, which is determined by the equilibrium between CN⁻ and HCN. This calculation is important in analytical chemistry and environmental science.

pH Calculation Formula

The pH of a NaCN solution can be calculated using the following steps:

1. Determine the concentration of HCN using the equilibrium constant for the reaction: CN⁻ + H₂O ⇌ HCN + OH⁻
2. Calculate the pOH using the concentration of OH⁻
3. Convert pOH to pH using the relationship pH + pOH = 14

Worked Example

Let's calculate the pH of a 0.082 M NaCN solution:

1. Assume the equilibrium constant K = 4.9 × 10⁻¹⁰ for the reaction CN⁻ + H₂O ⇌ HCN + OH⁻
2. Let x = [HCN] = [OH⁻] at equilibrium
3. Initial [CN⁻] = 0.082 M
4. At equilibrium: [CN⁻] = 0.082 - x
5. Set up the equilibrium expression: K = x² / (0.082 - x)
6. Solve for x: x² = 4.9 × 10⁻¹⁰(0.082 - x)
7. Approximate solution: x ≈ 7.0 × 10⁻⁶ M
8. Calculate pOH: pOH = -log[OH⁻] ≈ 5.15
9. Calculate pH: pH = 14 - pOH ≈ 8.85

Interpreting Results

A pH of approximately 8.85 indicates a slightly alkaline solution. This is expected because the CN⁻ ion reacts with water to produce OH⁻ ions. The solution is basic but not strongly basic.

In practical applications, this calculation helps determine the suitability of NaCN solutions for various chemical processes and environmental assessments.