Ka and Kb Relationship Equation, Derivation & Rearrangements

Core idea

Overview

This equation defines the inverse mathematical relationship between the dissociation constants of a conjugate acid-base pair in aqueous solution. It states that the product of the acid dissociation constant (Ka) and its conjugate base dissociation constant (Kb) is equal to the autoionization constant of water (Kw).

When to use: This formula is used when you need to find the strength of a conjugate base given the strength of its parent acid, or vice versa. It is only applicable to conjugate pairs in aqueous systems, typically at a standard temperature of 25°C where Kw is a known constant.

Why it matters: Understanding this relationship is essential for predicting the pH of salt solutions and the buffering capacity of chemical systems. It quantifies the principle that the conjugate of a strong acid is a weak base, which is a foundational concept in analytical chemistry and biochemistry.

Symbols

Variables

K_a = Acid Const Ka, K_b = Base Const Kb, K_w = Kw Constant

K_{a}

Acid Const Ka

V a r iab l e

K_{b}

Base Const Kb

V a r iab l e

K_{w}

Kw Constant

V a r iab l e

Walkthrough

Derivation

Understanding Ka and Kb Relationship

Relates Ka of a weak acid to Kb of its conjugate base via Kw.

Applies to conjugate acid-base pairs in aqueous solution.

1

State the Relationship:

For a conjugate pair, the product Ka×Kb equals Kw at that temperature.

K_{a} K_{b} = K_{w}

2

Log Form:

Taking -log10 of both sides gives the additive relationship.

p K_{a} + p K_{b} = p K_{w} (14 at 298 K)

Result

p K_{a} + p K_{b} = p K_{w} (14 at 298 K)

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

Free formulas

Rearrangements

Solve for $K_{a}$

Make Ka the subject

K_{a} = \frac{K _{w}}{K _{b}}

Start from the relationship between Ka, Kb, and Kw. To make Ka the subject, divide both sides by Kb.

Difficulty: 2/5

Solve for $K_{b}$

Make Kb the subject

K_{b} = \frac{K _{w}}{K _{a}}

Start from the relationship between the acid dissociation constant ( $K_{a}$ ), base dissociation constant ( $K_{b}$ ), and the ion-product constant for water ( $K_{w}$ ).

Difficulty: 2/5

Solve for $K_{w}$

Ka and Kb Relationship

K_{w} = K_{a} K_{b}

Start from the relationship between the acid dissociation constant ( $K_{a}$ ), the base dissociation constant ( $K_{b}$ ), and the ion product of water ( $K_{w}$ ). To make $K_{w}$ the subject, simply rearrange the terms.

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 hyperbola because Kb appears in the denominator of the relationship. As Kb increases, Ka decreases toward zero, and the curve approaches the axes as asymptotes, with the domain restricted to positive values. This shape illustrates an inverse relationship where a stronger base with a large Kb value must correspond to a weaker conjugate acid with a small Ka value. The most important feature is that the curve never reaches zero, meaning that even as one constant becomes extremely large, the other remain

Graph type: inverse

Why it behaves this way

Intuition

Imagine a fixed 'total strength' for a conjugate acid-base pair in water, represented by Kw; if the acid's strength (Ka) increases, its conjugate base's strength (Kb)

K_{a}

The acid dissociation constant, a quantitative measure of the strength of an acid in solution, indicating the extent to which it dissociates into its conjugate base and a proton

A larger Ka value signifies a stronger acid, meaning it releases more H+ ions into the solution.

K_{b}

The base dissociation constant, a quantitative measure of the strength of a base in solution, indicating the extent to which it dissociates into its conjugate acid and a hydroxide

A larger Kb value signifies a stronger base, meaning it produces more OH- ions in the solution.

K_{w}

The autoionization constant of water, which is the equilibrium constant for the self-ionization of water (2H2O ⇌ H3O+ + OH-), representing the product of the concentrations of

Kw is a fundamental constant for aqueous systems at a specific temperature (e.g., 1.0 x 10-14 at 25°C), establishing the inverse relationship between H3O+ and OH- concentrations in any aqueous solution.

Free study cues

Insight

Canonical usage

Equilibrium constants Ka and Kb are typically expressed in molarity (mol/L or M), resulting in Kw being expressed in $M^{2}$ .

Common confusion

A common mistake is forgetting that Kw is temperature-dependent and using the 25°C value for calculations at other temperatures, or incorrectly assigning units to Ka, Kb, or Kw.

Dimension note

While equilibrium constants (Ka, Kb, Kw) are rigorously dimensionless when expressed in terms of activities, in practical applications and at introductory levels, they are commonly reported with units derived from molar

Unit systems

$K_{a}$ M · Acid dissociation constant. In introductory contexts, it is often treated with units of molarity (mol/L) derived from concentration terms in the equilibrium expression.

$K_{b}$ M · Base dissociation constant. In introductory contexts, it is often treated with units of molarity (mol/L) derived from concentration terms in the equilibrium expression.

$K_{w}$ M^2 · Autoionization constant of water. Its value is highly temperature-dependent; the commonly cited value of 1.0 x 10^-14 M^2 is specific to 25°C.

One free problem

Practice Problem

A specific weak acid has a dissociation constant (Ka) of 1.8 × 10⁻⁵ at 25°C. Calculate the base dissociation constant (Kb) for its conjugate base.

Acid Const Ka0.000018

Kw Constant1e-14

Solve for: $K b$

Hint: Divide the water autoionization constant (Kw) by the given Ka value.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Finding Kb of acetate ion from Ka of acetic acid.

Study smarter

Tips

Always verify that the two species provided are a true conjugate pair differing by exactly one proton.
In most textbook problems, assume Kw is 1.0 × 10⁻¹⁴ unless a different temperature is specified.
If given pKa or pKb, convert them using 10 to the power of the negative value before using this multiplicative identity.

Avoid these traps

Common Mistakes

Using non-conjugate pairs.
Forgetting Kw changes with temperature.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

Relates Ka of a weak acid to Kb of its conjugate base via Kw.

This formula is used when you need to find the strength of a conjugate base given the strength of its parent acid, or vice versa. It is only applicable to conjugate pairs in aqueous systems, typically at a standard temperature of 25°C where Kw is a known constant.

Understanding this relationship is essential for predicting the pH of salt solutions and the buffering capacity of chemical systems. It quantifies the principle that the conjugate of a strong acid is a weak base, which is a foundational concept in analytical chemistry and biochemistry.

Using non-conjugate pairs. Forgetting Kw changes with temperature.

Finding Kb of acetate ion from Ka of acetic acid.

Always verify that the two species provided are a true conjugate pair differing by exactly one proton. In most textbook problems, assume Kw is 1.0 × 10⁻¹⁴ unless a different temperature is specified. If given pKa or pKb, convert them using 10 to the power of the negative value before using this multiplicative identity.

References

Sources

Atkins' Physical Chemistry
IUPAC Gold Book: acid dissociation constant
IUPAC Gold Book: base dissociation constant
IUPAC Gold Book: autoionization of water
Wikipedia: Conjugate acid-base pair
IUPAC Gold Book: 'acid dissociation constant, Ka'
IUPAC Gold Book: 'base dissociation constant, Kb'
IUPAC Gold Book: 'ionic product of water, Kw'

Ka and Kb Relationship

Overview

Variables

Derivation

State the Relationship:

Log Form:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Ionic Product of Water

Frequently Asked Questions

Sources