Calculate Oh and Ph for 0.10 M Nacn

Introduction

When calculating the pH and hydroxide ion concentration for a solution of sodium cyanide (NaCN), we need to account for the weak electrolyte nature of the compound. NaCN dissociates in water to form Na⁺ and CN⁻ ions, but the cyanide ion (CN⁻) reacts with water to form the cyanate ion (OCN⁻) and hydroxide ions (OH⁻).

This equilibrium reaction affects the pH of the solution, which we can calculate using the equilibrium constant for the reaction:

The equilibrium constant for this reaction is approximately 4.9 × 10⁻⁹ at 25°C.

Calculation Method

To calculate the pH and [OH⁻] for a 0.10 M NaCN solution, we follow these steps:

1. Assume the initial concentration of CN⁻ is equal to the concentration of NaCN (0.10 M).
2. Use the equilibrium expression to set up an equation for the change in concentration of CN⁻.
3. Solve for the equilibrium concentration of OH⁻.
4. Calculate the pH from the [OH⁻] using the relationship pH = 14 - pOH.

Example Calculation

Let's calculate the pH and [OH⁻] for a 0.10 M NaCN solution using the above method.

1. Initial [CN⁻] = 0.10 M
2. K_eq = 4.9 × 10⁻⁹
3. [OH⁻] = √(4.9 × 10⁻⁹ × 0.10) ≈ √(4.9 × 10⁻¹⁰) ≈ 7.0 × 10⁻⁶ M
4. pOH = -log(7.0 × 10⁻⁶) ≈ 5.15
5. pH = 14 - 5.15 ≈ 8.85

Therefore, for a 0.10 M NaCN solution, the hydroxide ion concentration is approximately 7.0 × 10⁻⁶ M and the pH is approximately 8.85.

Interpretation

A pH of 8.85 indicates that the solution is slightly basic. The hydroxide ion concentration of 7.0 × 10⁻⁶ M confirms this, as concentrations above 10⁻⁷ M are characteristic of basic solutions.

This calculation assumes ideal conditions and does not account for temperature effects or other potential interfering factors. For precise measurements, experimental verification is recommended.