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Acid dissociation constant

Expression for Ka of a weak acid.

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Core idea

Overview

The acid dissociation constant, Ka, is an equilibrium constant that measures the quantitative strength of an acid in an aqueous solution. It describes the ratio of the concentrations of the dissociated ions to the undissociated acid molecules at equilibrium.

When to use: This equation is used when calculating the extent of ionization for weak acids in water, as strong acids are assumed to dissociate completely. It is applicable in scenarios involving pH determination, buffer capacity calculations, and titration analysis in dilute solutions.

Why it matters: Ka values allow chemists to predict the acidity of a solution and the protonation state of molecules, which is vital in drug design and biochemistry. Understanding these constants is essential for controlling chemical reactions in industrial processes where pH must remain within a specific range.

Symbols

Variables

[H^+] = Hydrogen Ion, [A^-] = Conjugate Base, acid = Acid, K_a = Acid Dissociation Constant

Hydrogen Ion
Conjugate Base
Acid
Acid Dissociation Constant

Walkthrough

Derivation

Understanding Acid Dissociation Constant (Ka)

Quantifies the extent to which a weak acid dissociates in water; larger Ka means a stronger acid.

  • Acid is weak (partial dissociation).
1

State the Expression:

For HA \rightleftharpoons H^+ + A^-.

2

Common Weak-Acid Approximation:

If dissociation is small, $[HA]_{eq} \approx [HA]_{initial}$ and $[H^+]\approx[A^-]$.

Result

Source: Edexcel A-Level Chemistry — Acid-Base Equilibria

Free formulas

Rearrangements

Solve for

Make Ka the subject

Ka is already the subject of the formula.

Difficulty: 1/5

Solve for

Make [HA] the subject

Start with the acid dissociation constant formula. To make [HA] the subject, first multiply both sides by [HA] to clear the denominator, then divide both sides by K_a.

Difficulty: 2/5

Solve for

Make [H^+] the subject

Start from the acid dissociation constant. Multiply both sides by [HA] and then divide by [A^-] to isolate [H^+].

Difficulty: 2/5

Solve for

Make [A^-] the subject

Start from the acid dissociation constant formula. To make [A^-] the subject, first multiply by [HA], then divide by [H^+].

Difficulty: 2/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The graph is a straight line passing through the origin with a slope determined by the product of hydrogen and conjugate base concentrations. This linear relationship means that doubling the concentration of the acid directly doubles the product of the dissociated ions. For a chemistry student, this shows that as the acid concentration increases, the product of the ion concentrations must also increase proportionally to maintain a constant Ka. The most important feature is that the line passes through the origin, m

Graph type: linear

Why it behaves this way

Intuition

Imagine a dynamic balance where acid molecules are constantly breaking apart into ions and reforming, with K_a quantifying the preference for ions over intact molecules at equilibrium.

Acid dissociation constant
A larger K_a value means the acid dissociates more extensively, producing more H+ ions and resulting in a stronger acid.
Molar concentration of hydrogen ions (protons)
Directly indicates the acidity of the solution; a higher concentration means a lower pH and a more acidic solution.
Molar concentration of the conjugate base
Represents the amount of acid that has dissociated. It can also act as a proton acceptor.
Molar concentration of the undissociated weak acid
Represents the acid molecules that have not donated a proton. A higher concentration at equilibrium implies a weaker acid.

Signs and relationships

  • Denominator [HA]: The concentration of the undissociated acid is in the denominator because it represents the reactant in the dissociation equilibrium.
  • Numerator [H^+][A^-]: The product of the concentrations of the dissociated ions is in the numerator because they represent the products of the dissociation reaction.

Free study cues

Insight

Canonical usage

In introductory chemistry, K_a is typically treated as having units of concentration (e.g., mol/L or M) when molar concentrations are used in its calculation.

Common confusion

The primary confusion arises from whether K_a should have units or be dimensionless. This depends on whether concentrations are treated as activities (dimensionless) or molarities (with units of mol/L).

Dimension note

Although the acid dissociation constant K_a is strictly dimensionless when defined using activities, it is frequently treated as having units of concentration (e.g., mol/L or M)

Unit systems

mol/L · While rigorously dimensionless when defined using activities, K_a is commonly assigned units of concentration (e.g., mol/L or M) when calculated using molar concentrations in introductory contexts.
mol/L · Represents the molar concentration of hydrogen ions.
mol/L · Represents the molar concentration of the conjugate base.
mol/L · Represents the molar concentration of the undissociated weak acid.

Ballpark figures

  • Quantity:

One free problem

Practice Problem

A 0.10 M solution of a generic weak acid HA reaches equilibrium. The concentration of both H⁺ and A⁻ ions is measured to be 0.00134 M, and the remaining undissociated HA is 0.09866 M. Calculate the acid dissociation constant Ka.

Hydrogen Ion0.00134 mol/dm^3
Conjugate Base0.00134 mol/dm^3
Acid0.09866 mol/dm^3

Solve for: Ka

Hint: Square the ion concentration and divide by the concentration of the neutral acid.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Finding Ka of ethanoic acid from equilibrium data.

Study smarter

Tips

  • Larger Ka values correspond to stronger acids with higher degrees of ionization.
  • The concentration of liquid water is constant and therefore omitted from the expression.
  • At 25°C, pKa is calculated as the negative base-10 logarithm of Ka.

Avoid these traps

Common Mistakes

  • Using initial instead of equilibrium concentrations.
  • Forgetting the denominator.

Common questions

Frequently Asked Questions

Quantifies the extent to which a weak acid dissociates in water; larger Ka means a stronger acid.

This equation is used when calculating the extent of ionization for weak acids in water, as strong acids are assumed to dissociate completely. It is applicable in scenarios involving pH determination, buffer capacity calculations, and titration analysis in dilute solutions.

Ka values allow chemists to predict the acidity of a solution and the protonation state of molecules, which is vital in drug design and biochemistry. Understanding these constants is essential for controlling chemical reactions in industrial processes where pH must remain within a specific range.

Using initial instead of equilibrium concentrations. Forgetting the denominator.

Finding Ka of ethanoic acid from equilibrium data.

Larger Ka values correspond to stronger acids with higher degrees of ionization. The concentration of liquid water is constant and therefore omitted from the expression. At 25°C, pKa is calculated as the negative base-10 logarithm of Ka.

References

Sources

  1. IUPAC Gold Book: Acid dissociation constant
  2. Atkins' Physical Chemistry
  3. Wikipedia: Acid dissociation constant
  4. IUPAC Gold Book: 'acid dissociation constant'
  5. General Chemistry: Principles and Modern Applications by Petrucci, Herring, Madura, Bissonnette
  6. Atkins' Physical Chemistry, 11th Edition
  7. Edexcel A-Level Chemistry — Acid-Base Equilibria