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Henderson-Hasselbalch equation

Calculate pH of a buffer solution.

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p H = p K_{a} + lo g_{10} \frac{[ A ^{-} ]}{[ H A ]}

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Core idea

Overview

The Henderson-Hasselbalch equation describes the relationship between the pH of a buffer solution and the equilibrium concentrations of a weak acid and its conjugate base. It is a logarithmic transformation of the acid dissociation constant expression, facilitating the calculation of acidity in complex biological and chemical systems.

When to use: Apply this equation when calculating the pH of a buffer solution consisting of a weak acid and its salt. It is most reliable when the ratio of acid to base is between 0.1 and 10 and the concentrations are high enough to ignore the self-ionization of water.

Why it matters: This formula is essential for understanding physiological buffering, such as the bicarbonate system that regulates human blood pH. It also enables scientists to design stable environments for enzyme assays and industrial fermentation processes.

Symbols

Variables

pK_a = Acid Dissociation Constant, [A^-] = Conjugate Base, [HA] = Acid, pH = Potential of Hydrogen

p K_{a}

Acid Dissociation Constant

V a r iab l e

[A^{-}]

Conjugate Base

m o l / d m^{3}

[H A]

Acid

m o l / d m^{3}

p H

Potential of Hydrogen

V a r iab l e

Walkthrough

Derivation

Derivation of the Henderson-Hasselbalch Equation

Relates buffer pH to pKa and the ratio of conjugate base to acid concentrations.

Weak acid dissociates only slightly.
[HA]_{eq} \approx [HA]_{initial} and [A^-]_{eq} \approx [A^-]_{initial} (buffer approximation).

Start with the Ka Expression:

Equilibrium expression for HA \rightleftharpoons H^+ + A^-.

K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]}

Rearrange for [H+]:

Make hydrogen ion concentration the subject.

[H^{+}] = K_{a} \frac{[ H A ]}{[ A ^{-} ]}

Take -log10:

Convert to pH and pKa using logarithm rules.

- lo g_{10} [H^{+}] = - lo g_{10} K_{a} - lo g_{10} (\frac{[ H A ]}{[ A ^{-} ]})

Write the Final Form:

This is the Henderson–Hasselbalch equation used for buffer calculations.

pH = p K_{a} + lo g_{10} (\frac{[ A ^{-} ]}{[ H A ]})

Result

pH = p K_{a} + lo g_{10} (\frac{[ A ^{-} ]}{[ H A ]})

Source: OCR A-Level Chemistry A — pH and Buffers

Free formulas

Rearrangements

Solve for $p H$

Make pH the subject

p H = p K_{a} + lo g_{10} \frac{[ A ^{-} ]}{[ H A ]}

pH is already the subject of the formula.

Difficulty: 1/5

Solve for $p K_{a}$

Make pKa the subject

p K_{a} = p H - lo g_{10} \frac{[ A ^{-} ]}{[ H A ]}

Rearrange the Henderson-Hasselbalch equation to isolate the acid dissociation constant, pK_a, by subtracting the logarithmic term from both sides.

Difficulty: 2/5

Solve for $[A^{-}]$

Make [A^-] the subject

[A^{-}] = [H A] \cdot 1 0^{(p H - p K_{a})}

To make the concentration of the conjugate base, $[A^{-}]$ , the subject of the Henderson-Hasselbalch equation, first isolate the base-10 logarithm term, then apply the inverse exponential function, and finally multiply to isolate $[A^{-}]$ .

Difficulty: 2/5

Solve for $[H A]$

Make [HA] the subject

[H A] = \frac{[ A ^{-} ]}{1 0 ^{(p H - p K_{a})}}

Start from the Henderson-Hasselbalch equation. To make [HA] the subject, first isolate the base-10 logarithm term by subtracting $p K_{a}$ , then raise 10 to the power of both sides to remove the logarithm, and finally rearrange the equation to...

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 with a slope of one, showing that pKa and pH maintain a direct, proportional relationship. For a chemistry student, this means that a larger pKa value corresponds to a higher pH, indicating a weaker acid component in the buffer system. The most important feature is that the linear relationship means any change in the pKa value results in an identical shift in the pH, demonstrating that the two variables are perfectly linked in a one-to-one ratio.

Graph type: linear

Why it behaves this way

Intuition

Imagine a chemical seesaw where the pK_a is the pivot point. The pH of the buffer solution is determined by the balance between the weak acid [HA] on one side and its conjugate base [A^-] on the other, with the logarithm

p H

A measure of the hydrogen ion activity (effective concentration) in an aqueous solution.

Indicates acidity or basicity; lower values are more acidic, higher values are more basic.

p K_{a}

The negative base-10 logarithm of the acid dissociation constant (Ka) for a weak acid.

A characteristic constant for a given weak acid, indicating its strength; smaller pKa means a stronger weak acid.

[A^{-}]

The molar concentration of the conjugate base of the weak acid in the solution.

Represents the base component of the buffer, which can accept protons (H+).

[H A]

The molar concentration of the weak acid in the solution.

Represents the acid component of the buffer, which can donate protons (H+).

Signs and relationships

+log_{10}\frac{[A^-]}{[HA]}: The positive sign indicates that as the ratio of conjugate base [A^-] to weak acid [HA] increases, the pH of the solution also increases, making it more basic.
log_{10}\frac{[A^-]}{[HA]}: The logarithmic function converts the ratio of concentrations into a linear scale, making the relationship with pH (which is also a logarithmic scale) straightforward.

Free study cues

Insight

Canonical usage

The Henderson-Hasselbalch equation relates dimensionless quantities (pH, pK_a) to a dimensionless ratio of concentrations, requiring consistent units for the acid and base concentrations.

Common confusion

A common mistake is using different concentration units for the conjugate base and weak acid, which would lead to an incorrect dimensionless ratio, or attempting to assign units to pH or pK_a.

Dimension note

All terms in the Henderson-Hasselbalch equation (pH, pK_a, and the ratio of conjugate base to weak acid concentrations) are dimensionless quantities.

Unit systems

$[A^{-}]$ mol/L · Represents the equilibrium concentration of the conjugate base. Must be in the same concentration units as [HA] for the ratio to be dimensionless.

$[H A]$ mol/L · Represents the equilibrium concentration of the weak acid. Must be in the same concentration units as [A-] for the ratio to be dimensionless.

$p H$ dimensionless · Defined as -log10[H+], where [H+] is the numerical value of the hydrogen ion activity or concentration in mol/L.

$p K_{a}$ dimensionless · Defined as -log10(K_a), where K_a is the acid dissociation constant.

Ballpark figures

Quantity:
Quantity:

One free problem

Practice Problem

An ethanoic acid / sodium ethanoate buffer is prepared. pKa of ethanoic acid = 4.76. [A-] = 0.08 mol/dm^3 and [HA] = 0.02 mol/dm^3. Calculate the pH.

Acid Dissociation Constant4.76

Conjugate Base0.08 mol/dm^3

Acid0.02 mol/dm^3

Solve for: pH

Hint: pH = pKa + log10([A-]/[HA]).

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Calculating pH of an ethanoic acid / sodium ethanoate buffer.

Study smarter

Tips

Ensure the concentrations of [A⁻] and [HA] use the same molar units.
pH equals pKa exactly when the concentrations of the acid and conjugate base are equal.
The equation becomes less accurate at extreme pH values where the acid or base is very dilute.

Avoid these traps

Common Mistakes

Swapping acid and base in the ratio.
Using ln instead of log10.

Common questions

Frequently Asked Questions

Relates buffer pH to pKa and the ratio of conjugate base to acid concentrations.

Apply this equation when calculating the pH of a buffer solution consisting of a weak acid and its salt. It is most reliable when the ratio of acid to base is between 0.1 and 10 and the concentrations are high enough to ignore the self-ionization of water.

This formula is essential for understanding physiological buffering, such as the bicarbonate system that regulates human blood pH. It also enables scientists to design stable environments for enzyme assays and industrial fermentation processes.

Swapping acid and base in the ratio. Using ln instead of log10.

Calculating pH of an ethanoic acid / sodium ethanoate buffer.

Ensure the concentrations of [A⁻] and [HA] use the same molar units. pH equals pKa exactly when the concentrations of the acid and conjugate base are equal. The equation becomes less accurate at extreme pH values where the acid or base is very dilute.

References

Sources

Atkins' Physical Chemistry
IUPAC Gold Book
Wikipedia: Henderson-Hasselbalch equation
IUPAC Gold Book: pH
IUPAC Gold Book: acid dissociation constant, K_a
McQuarrie, Donald A. General Chemistry
Chemistry: The Central Science by Brown, LeMay, Bursten, Murphy, Woodward, and Stoltzfus
Analytical Chemistry by Gary D. Christian, Purnendu K. Dasgupta, and Kevin A. Schug

Henderson-Hasselbalch equation

Overview

Variables

Derivation

Start with the Ka Expression:

Rearrange for [H+]:

Take -log10:

Write the Final Form:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Acid dissociation constant

Frequently Asked Questions

Sources