Dissolving Sodium Acetate in Water: Determining Hydroxide Ion Concentration

Sodium acetate, a common salt used in a variety of applications, dissociates in water to form sodium ions and acetate ions. When dissolved in water, these acetate ions can undergo a hydrolysis reaction, leading to the production of hydrogen ions and hydroxide ions. In this article, we will explore the process of determining the hydroxide ion concentration in a solution prepared by dissolving 1 mole of sodium acetate in 1 liter of water.

Introduction to Sodium Acetate

Sodium acetate, or sodium ethanoate (NaCH3COO), is a salt derived from acetic acid (CH3COOH) and sodium hydroxide (NaOH). Its dissociation in water follows the equation:

Sodium acetate: NaCH3COO rarr; Na CH3COO-

Hydrolysis Reaction: Acetate Ion and Water Interaction

Acetate ions (CH3COO-) can react with water to form acetic acid and hydroxide ions in a process known as hydrolysis:

CH3COO- H2O leftrightarrow; CH3COOH OH-

This reaction establishes a dynamic equilibrium where products and reactants coexist, and it is crucial for determining the hydroxide ion concentration in the solution.

Determining the Equilibrium Constant (Kb)

To calculate the hydroxide ion concentration, we start by determining the equilibrium constant (Kb) for the hydrolysis reaction. Kb is related to the acid dissociation constant (Ka) of acetic acid and the ion product of water (Kw):

Kw Ka × Kb

Given:

Kw 1.0 × 10-14 (at 25°C)

Ka for acetic acid 1.8 × 10-5

Calculating Kb:

Kb Kw / Ka (1.0 × 10-14) / (1.8 × 10-5) ≈ 5.56 × 10-10

Setting Up the Equilibrium Expression

Let's assume the initial concentration of acetate ions (CH3COO-) is 1 M. At equilibrium, we establish the following concentrations:

[CH3COO-] 1 - x [CH3COOH] x [OH-] x

The equilibrium expression for Kb is given by:

Kb [CH3COOH][OH-] / [CH3COO-]

Substituting the values:

Kb x × x / (1 - x)

Since Kb is very small, we can approximate (1 - x) ≈ 1:

Kb ≈ x2 / 1 x2

Solving for Hydroxide Ion Concentration

To solve for x:

x2 Kb 5.56 × 10-10

Taking the square root of both sides:

x  √(5.56 × 10-10) ≈ 2.36 × 10-5 M

Thus, the concentration of hydroxide ions (OH-) in the solution is approximately 2.36 × 10-5 M.

Conclusion

The determination of the hydroxide ion concentration in a sodium acetate solution involves understanding the hydrolysis reaction and applying the equilibrium constants. By following the steps outlined above, we can accurately calculate the concentration of hydroxide ions (OH-) in the solution, crucial for a variety of chemical and biological applications.

Keywords: Sodium acetate, hydroxide ion concentration, hydrolysis reaction