Chemistry

Acid Dissociation Constant

Estimate K_a from the equilibrium composition of a dilute monoprotic acid.

K_{a} ≃ \frac{[ H _{3} O ^{+} ] [ A ^{-} ]}{[ HA ] c ^{\circ}}

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Difficulty

Intermediate

Read time

4 min

Prerequisites

Algebra, logarithms, and molar quantities

Review statusNeeds review

Source

OpenStax, Chemistry 2e, §14.3

Adapted from Teng-Jui Lin’s CHEM 155 equation sheet. The linked textbook reference was used to check the equation and its physical context. Pending editorial review.

Adapted from Teng-Jui Lin, Equation Sheets, under CC BY 4.0.

FormuLab adapted notation and chemical symbols, added variable meanings, units, applicability conditions, and an original worked example. Restored concentration brackets and made standard-state normalization explicit.

Definition

Understanding the core concept

For HA donating one proton to water, K_a compares the product activities with the undissociated-acid activity. The concentration form shown normalizes by the standard concentration and approximates activity coefficients as one.

Variables & Units

Understanding each component

Symbol	Meaning	Units
$K_{a}$	Dimensionless acid dissociation constant	$1$
$[H_{3} O^{+}]$	Equilibrium hydronium concentration	$mol L^{- 1}$
$[A^{-}]$	Equilibrium conjugate-base concentration	$mol L^{- 1}$
$[HA]$	Equilibrium undissociated-acid concentration	$mol L^{- 1}$
$c^{\circ}$	Standard concentration, 1 mol/L	$mol L^{- 1}$

Real-World Applications

Where this formula is used in practice

Weak acids

Relate equilibrium composition to acid strength.

Equilibrium data

Estimate a dissociation constant from concentration measurements.

Worked Example

Step-by-step calculation with real numbers

Problem

A dilute monoprotic-acid solution at equilibrium has [H₃O⁺] = [A⁻] = 0.00100 mol/L and [HA] = 0.100 mol/L. Estimate dimensionless K_a.

Given

[H_{3} O^{+}] = 0.00100 mol L^{- 1}

[A^{-}] = 0.00100 mol L^{- 1}

[HA] = 0.100 mol L^{- 1}

c^{\circ} = 1 mol L^{- 1}

Solution

K_{a} ≃ \frac{[ H _{3} O ^{+} ] [ A ^{-} ]}{[ HA ] c ^{\circ}}

K_{a} ≃ \frac{( 0.00100 ) ( 0.00100 )}{( 0.100 ) ( 1 )}

K_{a} \approx 1 \times 1 0^{- 5}

Final Answer

K_{a} \approx 1 \times 1 0^{- 5}

Formula Guide

Info

DifficultyIntermediate

Read Time4 min

PrerequisitesAlgebra, logarithms, and molar quantities

Tools

Source & review

StatusNeeds review

Last reviewedPending

Source

OpenStax, Chemistry 2e, §14.3

Adapted from Teng-Jui Lin’s CHEM 155 equation sheet. The linked textbook reference was used to check the equation and its physical context. Pending editorial review.

Adapted from Teng-Jui Lin, Equation Sheets, under CC BY 4.0.

Assumptions

• Use equilibrium concentrations for the monoprotic reaction HA + H₂O ⇌ H₃O⁺ + A⁻.
• Assume a dilute aqueous solution, approximately unit water activity, and negligible activity-coefficient corrections.

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