Calculate The Ph of A 0.082 M Solution of Nacn

Sodium cyanide (NaCN) is a strong electrolyte that dissociates completely in water. Calculating the pH of a 0.082 M solution of NaCN requires understanding the dissociation of NaCN and the resulting equilibrium concentrations of CN⁻ and H⁺ ions.

Introduction

Sodium cyanide (NaCN) is a highly toxic compound that dissociates completely in water according to the following equation:

NaCN (s) → Na⁺ (aq) + CN⁻ (aq)

The cyanide ion (CN⁻) reacts with water to form the cyanate ion (OCN⁻) and hydrogen ions (H⁺):

CN⁻ (aq) + H₂O (l) ⇌ OCN⁻ (aq) + H⁺ (aq)

This reaction is the basis for calculating the pH of a NaCN solution. The equilibrium constant for this reaction is approximately 4.7 × 10⁻⁹ at 25°C.

How to Calculate the pH

To calculate the pH of a 0.082 M NaCN solution, follow these steps:

Determine the initial concentration of CN⁻ ions. Since NaCN dissociates completely, [CN⁻] = [NaCN] = 0.082 M.
Set up the equilibrium expression for the reaction between CN⁻ and H₂O:

K = [OCN⁻][H⁺] / [CN⁻]

Assume that x is the concentration of H⁺ ions formed. Since the reaction consumes CN⁻ and produces OCN⁻ and H⁺, the equilibrium concentrations are:

[CN⁻] = 0.082 - x [OCN⁻] = x [H⁺] = x

Substitute these into the equilibrium expression:

4.7 × 10⁻⁹ = (x)(x) / (0.082 - x)

Solve the quadratic equation to find x.
Calculate the pH using the concentration of H⁺ ions: pH = -log[H⁺].

Example Calculation

Let's calculate the pH of a 0.082 M NaCN solution step by step.

Step 1: Initial Setup

Given:

Initial [CN⁻] = 0.082 M
K = 4.7 × 10⁻⁹

Step 2: Equilibrium Expression

4.7 × 10⁻⁹ = x² / (0.082 - x)

Step 3: Solve the Quadratic Equation

Rearrange the equation:

x² + 4.7 × 10⁻⁹x - 3.878 × 10⁻¹⁰ = 0

Using the quadratic formula:

x = [-b ± √(b² - 4ac)] / 2a

Where a = 1, b = 4.7 × 10⁻⁹, and c = -3.878 × 10⁻¹⁰.

Calculate the discriminant:

D = (4.7 × 10⁻⁹)² - 4(1)(-3.878 × 10⁻¹⁰) = 2.209 × 10⁻¹⁷ + 1.551 × 10⁻⁹ ≈ 1.551 × 10⁻⁹

Solve for x:

x = [-4.7 × 10⁻⁹ ± √(1.551 × 10⁻⁹)] / 2

We take the positive root since concentration cannot be negative:

x ≈ [ -4.7 × 10⁻⁹ + 3.938 × 10⁻⁵ ] / 2 ≈ 1.969 × 10⁻⁵ M

Step 4: Calculate pH

pH = -log[H⁺] = -log(1.969 × 10⁻⁵) ≈ 4.70

The pH of a 0.082 M NaCN solution is approximately 4.70.

Interpreting the Results

A pH of 4.70 indicates that the solution is acidic. This is expected because the reaction between CN⁻ and H₂O produces H⁺ ions. The low pH is characteristic of cyanide solutions and is important in industrial applications where cyanide is used in gold extraction and metal plating.

Safety Note: Sodium cyanide is highly toxic. Proper safety precautions must be taken when handling cyanide solutions.

Frequently Asked Questions

What is the pH of a 0.082 M NaCN solution?: The pH of a 0.082 M NaCN solution is approximately 4.70.
Why is the pH of NaCN solutions acidic?: The pH is acidic because the CN⁻ ion reacts with water to form H⁺ ions.
What is the equilibrium constant for the reaction between CN⁻ and H₂O?: The equilibrium constant is approximately 4.7 × 10⁻⁹ at 25°C.
How does the concentration of NaCN affect the pH?: Higher concentrations of NaCN will produce more H⁺ ions, resulting in a lower pH.
Is NaCN a strong or weak electrolyte?: NaCN is a strong electrolyte because it dissociates completely in water.