9.4 Ionisation Constant of Water and Calculation of pH

This topic explains the ionisation constant of water (K_w) and how pH is calculated. It covers the autoionization of water, the relationship between [H⁺] and [OH^-], and the pH scale.

Definition

Ionisation Constant of Water (K_w): The product of the concentrations of hydrogen ions and hydroxide ions in water at equilibrium.

K_w = [H⁺][OH^-]

pH: The negative logarithm of the hydrogen ion concentration.

pH = -log[H⁺]

Main Idea

Water is a unique compound due to its ability to accept or donate protons under different environments. It acts as a Brønsted acid in the presence of ammonia and as a Brønsted base in the presence of hydrochloric acid. In fact, water itself undergoes ionization to a small extent. The equilibrium constant for this autoionization is K_w. Since the concentrations of [H⁺] and [OH^-] are usually very small, a more practical measure called pH was proposed.

Core Concept

• Autoionization of water: H₂O + H₂O ⇌ H₃O⁺ + OH^-
• K_w = [H⁺][OH^-] = 1.0 × 10^-14 at 25°C
• pH = -log[H⁺]
• Neutral solution: [H⁺] = [OH^-] = 1.0 × 10^-7 M, pH = 7
• Acidic solution: [H⁺] > 1.0 × 10^-7 M, pH < 7
• Basic solution: [H⁺] < 1.0 × 10^-7 M, pH > 7
• pH + pOH = 14

Detailed Explanation

Autoionization of Water

Water itself undergoes ionization to a small extent as shown in the following equation:

H₂O + H₂O ⇌ H₃O⁺ + OH^-

This reaction is regarded as autoionization of water. There is an equilibrium between water molecules (on the left side of the equation) and the hydronium ions and hydroxide ions (on the right side of the equation).

Equilibrium Constant for Water

The equilibrium constant (K) for this equilibrium can be expressed by the following equation:

K = [H₃O⁺][OH^-][H₂O]²

Since water is a weak electrolyte and solvent, the concentration of water, [H₂O], is in large excess, therefore it remains constant. The above equation may be re-written as:

K[H₂O]² = [H₃O⁺][OH^-]

The term K[H₂O]² is the product of two constants and is represented by K_w:

K_w = [H₃O⁺][OH^-]

K_w is termed as the ionic product constant of water. Since [H₃O⁺] is the concentration of hydrated protons at equilibrium, the above equation corresponds to:

K_w = [H⁺][OH^-]

In pure water at 25°C:

K_w = [H⁺][OH^-] = 1.0 × 10^-14

Whenever [H⁺] = [OH^-], the aqueous solution is found to be neutral, neither acidic nor basic.

[H⁺] = 1.0 × 10^-7 M \] \[ [OH^-] = 1.0 × 10^-7 M \] \[ [H⁺][OH^-] = [H⁺]² = 1.0 × 10^-14

It is to be noted that, because K_w is an equilibrium constant, it is temperature dependent. Thus, at 40°C, K_w = 3.8 × 10^-14, which corresponds to:

[H⁺] = 1.9 × 10^-7 M [OH^-] = 1.9 × 10^-7 M as [H⁺] = [OH^-].

pH

Since the concentrations of [H⁺] and [OH^-] are usually very small numbers and inconvenient to work with, a more practical measure called pH was proposed and defined as:

pH = -log[H⁺]

It means that pH of a solution is given by the negative logarithm of the [H⁺] concentration (in mol/dm³). However, it must be kept in mind that pH being a logarithmic value, does not have any units.

The pH concept implies that at 25°C, the different types of solutions will show the following behaviours:

Acidic solution: [H⁺] > 1.0 × 10^-7 M, pH < 7.00

Basic solution: [H⁺] < 1.0 × 10^-7 M, pH > 7.00

Neutral solution: [H⁺] = 1.0 × 10^-7 M, pH = 7.00

Relationship Between pH and pOH

As K_w = [H⁺][OH^-]

At 25°C, K_w = 1.0 × 10^-14

1.0 × 10^-14 = [H⁺][OH^-]

Taking log of both sides:

log(1.0 × 10^-14) = log[H⁺] + log[OH^-]

-14 = -log[H⁺] - log[OH^-]

14 = (-log[H⁺]) + (-log[OH^-])

14 = pH + pOH

Example 9.2: Calculating [H⁺] from [OH^-]

The concentration of [OH^-] ion in a household ammonia solution is 0.005 M. Calculate the concentration of [H⁺] in it.

Solution:

[OH^-] = 0.005 M

K_w = [H⁺][OH^-]

1.0 × 10^-14 = [H⁺] × 0.005

[H⁺] = 1.0 × 10^-140.005 = 2.0 × 10^-12 M

Example 9.3: Calculating pH of HCl Solution

Calculate the pH of 0.001 M aqueous hydrochloric acid solution.

Solution:

Hydrochloric acid ionizes in water completely; therefore,

HCl + H₂O → H₃O⁺ + Cl^-

0.001 M 0.001 M

[H₃O⁺]\) is in fact the same as \([H⁺]

Therefore [H⁺] = 0.001 M

pH = -log(0.001) \] \[ = -log 10^-3 \] \[ = 3.00

Therefore, the pH of 0.001 M aqueous hydrochloric acid is 3.00.

Example 9.4: Calculating pH of NaOH Solution

Calculate the pH of 0.062 M NaOH solution.

Solution:

NaOH_(aq) → Na⁺_(aq) + OH^-_(aq)

0.062 M 0.062 M

[OH^-] = 0.062 M

pOH = -log[OH^-] = -log(0.062) = 1.21

pH = 14 - pOH = 14 - 1.21 = 12.79

Intermediate Level

K_w and pH

Types of Solutions

Examples

Example 9.2: [OH^-] = 0.005 M [H⁺] = 1.0 × 10^-140.005 = 2.0 × 10^-12 M

Example 9.3: [H⁺] = 0.001 M pH = 3.00

Example 9.4: [OH^-] = 0.062 M pOH = 1.21 pH = 12.79

Important Points

• 1. K_w = [H⁺][OH^-] = 1.0 × 10^-14 at 25°C.
• 2. pH = -log[H⁺].
• 3. pOH = -log[OH^-].
• 4. pH + pOH = 14.
• 5. Acidic: pH < 7; Neutral: pH = 7; Basic: pH > 7.
• 6. Example 9.2: [H⁺] = 2.0 × 10^-12 M.
• 7. Example 9.3: pH = 3.00.
• 8. Example 9.4: pH = 12.79.

Remember

• - K_w: [H⁺][OH^-] = 1.0 × 10^-14
• - pH: -log[H⁺]
• - pOH: -log[OH^-]
• - **pH + pOH = 14**
• - Acidic: pH < 7
• - **Neutral:** pH = 7
• - Basic: pH > 7
• - **Examples:** pH = 3.00 (HCl), pH = 12.79 (NaOH)

Common Mistakes

• 1. Forgetting that K_w is temperature dependent: It changes with temperature.
• 2. Mixing up pH and pOH: pH = -log[H⁺]; pOH = -log[OH^-].
• 3. **Forgetting the relationship:** pH + pOH = 14.
• 4. Thinking pH has units: pH is dimensionless.

Quick Revision

• K_w = [H⁺][OH^-] = 1.0 × 10^-14 at 25°C.
• pH = -log[H⁺].
• pOH = -log[OH^-].
• pH + pOH = 14.
• Acidic: pH < 7; Neutral: pH = 7; Basic: pH > 7.
• Example 9.2: [H⁺] = 2.0 × 10^-12 M.
• Example 9.3: pH = 3.00.
• Example 9.4: pH = 12.79.

60-Second Revision

• K_w: [H⁺][OH^-] = 1.0 × 10^-14
• pH: -log[H⁺]
• pOH: -log[OH^-]
• **pH + pOH = 14**
• Acidic: pH < 7
• **Neutral:** pH = 7
• Basic: pH > 7
• **Examples:** pH = 3.00 (HCl), pH = 12.79 (NaOH)

QUALITY VALIDATION

• Topic: 9.4 Ionisation Constant of Water and Calculation of pH
• Chapter: 9 (Acids-Bases Chemistry)

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