Ka and weak acids

What you'll learn

What “weak acid” really means, and how it differs from “dilute”.
How to write and use the acid dissociation constant, $K_a$ .
How to calculate the pH of a weak acid from $K_a$ and concentration.
How to calculate $K_a$ from a measured pH.

Prerequisites: acids, hydrogen ions and pH

A Brønsted–Lowry acid is a substance that donates a proton, H⁺. In aqueous solution, H⁺ is more accurately present as hydroxonium ions, H₃O⁺, but A-Level calculations usually use the shorthand H⁺.

The pH scale measures hydrogen ion concentration. Lower pH means higher concentration of H⁺ ions.

Definition

For an aqueous solution,

pH=−log⁡10[H+]\text{pH} = -\log_{10}[H^+]

where $H^+]$ is the hydrogen ion concentration in mol dm⁻³.

You can rearrange this to find hydrogen ion concentration:

[H+]=10−pH[H^+] = 10^{-\text{pH}}

Example

Converting pH to hydrogen ion concentration

A solution has pH 3.40. Find $H^+]$ .

Use the rearranged pH equation:
$[H+]=10−pH[H^+] = 10^{-\text{pH}}$
Substitute pH 3.40:
$H^+] = 10^{-3.40}$
Calculate the concentration:
$dm−3[H^+] = 3.98 \times 10^{-4}\ \text{mol dm}^{-3}$

Strong, weak, concentrated and dilute

The words strong and weak describe how much an acid dissociates into ions in water.

A strong acid fully dissociates in aqueous solution. For example:

HCl(aq) → H⁺(aq) + Cl⁻(aq)

A weak acid only partially dissociates, so an equilibrium is set up. For a general monoprotic acid, HA:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

A monoprotic acid donates one proton per acid molecule. The species A⁻ is the conjugate base, formed after HA has donated H⁺.

Definition

Weak acid

A weak acid is an acid that partially dissociates in aqueous solution, forming an equilibrium mixture of undissociated acid molecules and ions.

At equilibrium, a weak acid solution usually contains mostly HA molecules and relatively few H⁺ and A⁻ ions. This is the key particle-level idea behind weak-acid pH calculations.

Particle diagram showing a weak acid HA partially dissociating into H+ and A- at equilibrium, with the Ka expression

Key Idea

Strength is not concentration

Strong/weak tells you the extent of dissociation. Concentrated/dilute tells you how much acid is dissolved per dm³. A dilute acid can still be strong, and a concentrated acid can still be weak.

The acid dissociation constant, Ka

For the weak monoprotic acid HA:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

the equilibrium constant is called the acid dissociation constant, written $K_a$ .

Definition

Acid dissociation constant

For a weak acid HA,

Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}

using equilibrium concentrations. A larger $K_a$ means the acid is more dissociated, so it is a stronger weak acid.

For this expression, $K_a$ usually has units of mol dm⁻³ at A-Level, because the numerator has two concentration terms and the denominator has one.

$K_a$ is temperature-dependent, so values should only be compared at the same temperature.

pKa

You may also see pKa, defined as:

pKa=−log⁡10Ka\text{p}K_a = -\log_{10}K_a

A lower pKa means a stronger acid, because it corresponds to a larger $K_a$ .

Example

Writing a Ka expression

Methanoic acid, HCOOH, partially dissociates:

HCOOH(aq) ⇌ H⁺(aq) + HCOO⁻(aq)

Identify the weak acid as HCOOH and the ions produced as H⁺ and HCOO⁻.
Put the product concentrations on the top and the undissociated acid concentration on the bottom:
$Ka=[H+][HCOO−][HCOOH]K_a = \frac{[H^+][\text{HCOO}^-]}{[\text{HCOOH}]}$
Use equilibrium concentrations only; do not use the initial acid concentration unless an approximation has been justified.

Calculating pH of a weak acid from Ka

Suppose you have a weak monoprotic acid HA with initial concentration $c$ .

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

Let the equilibrium concentration of H⁺ be $x$ .

Because the acid is monoprotic:

$H^+] = x$
$A^-] = x$
$[H A] = c - x$

So:

Ka=x2c−xK_a = \frac{x^2}{c - x}

For most weak acid calculations at A-Level, the acid is only slightly dissociated, so $x$ is much smaller than $c$ . That means:

\approx c

So the expression simplifies to:

Ka≈x2cK_a \approx \frac{x^2}{c}

and therefore:

[H+]=x≈Kac[H^+] = x \approx \sqrt{K_a c}

Example

Calculating the pH of a weak acid

Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, CH₃COOH, with $dm−3K_a = 1.74 \times 10^{-5}\ \text{mol dm}^{-3}$ .

Write the dissociation equation and link the concentrations:

CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq)

Since one molecule of acid forms one H⁺ and one CH₃COO⁻, let $H^+] = [CH_3COO^-] = x$ .
Use the weak-acid approximation:
$Ka≈x2cK_a \approx \frac{x^2}{c}$
so:
$\approx \sqrt{K_a c}$
Substitute the values:
$\sqrt{(1.74 \times 10^{-5})(0.100)}$ $\times 10^{-3}\ \text{mol dm}^{-3}$
Convert $H^+]$ to pH:
$pH=−log⁡10(1.32×10−3)\text{pH} = -\log_{10}(1.32 \times 10^{-3})$ $pH=2.88\text{pH} = 2.88$
Check the approximation:
$xc×100=1.32×10−30.100×100=1.32%\frac{x}{c} \times 100 = \frac{1.32 \times 10^{-3}}{0.100} \times 100 = 1.32\%$
This is small, so the approximation is reasonable.

Common Mistake

Treating a weak acid like a strong acid

For a 0.100 mol dm⁻³ weak acid, $H^+]$ is not 0.100 mol dm⁻³. Only a small fraction of the acid molecules dissociate, so you must use $K_a$ to find $H^+]$ .

Calculating Ka from pH

Sometimes you are given the pH of a weak acid solution and asked to find $K_a$ .

The usual method is:

Convert pH to $H^+]$ .
Use the 1:1 ratio to find $A^-]$ .
Work out the remaining $[H A]$ .
Substitute into the $K_a$ expression.

For a weak monoprotic acid with no other source of H⁺:

H^+] = [A^-]

and:

HA] = c - [H^+]

Example

Calculating Ka from pH

A 0.0500 mol dm⁻³ solution of a weak monoprotic acid HA has pH 2.93. Calculate $K_a$ .

Convert pH to hydrogen ion concentration:
$H^+] = 10^{-2.93}$ $dm−3[H^+] = 1.17 \times 10^{-3}\ \text{mol dm}^{-3}$
Use the dissociation ratio:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

Since the ratio is 1:1:
$dm−3[A^-] = 1.17 \times 10^{-3}\ \text{mol dm}^{-3}$
Find the equilibrium concentration of undissociated acid:
$\times 10^{-3}$ $0.0488\ \text{mol dm}^{-3}$
Substitute into the $K_a$ expression:
$Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}$ $Ka=(1.17×10−3)(1.17×10−3)0.0488K_a = \frac{(1.17 \times 10^{-3})(1.17 \times 10^{-3})}{0.0488}$ $dm−3K_a = 2.81 \times 10^{-5}\ \text{mol dm}^{-3}$

The weak-acid approximation

The approximation $\approx c$ works when the acid dissociates only slightly. A common check is whether the calculated dissociation is less than about 5% of the original concentration.

If:

xc×100\frac{x}{c} \times 100

is small, the approximation is acceptable.

Common Mistake

When the approximation breaks

If the calculated $H^+]$ is not small compared with the original acid concentration, do not use $\approx c$ . Use $Ka=x2c−xK_a = \frac{x^2}{c - x}$ and solve the resulting quadratic equation.

Dilution and weak acids

For a weak acid where the approximation is valid:

[H+]≈Kac[H^+] \approx \sqrt{K_a c}

This means $H^+]$ depends on the square root of concentration.

So if you dilute a weak acid by a factor of 10, the hydrogen ion concentration falls by about $10\sqrt{10}$ , not by 10. The pH therefore increases by about 0.50, not 1.00.

Tip

Dilution sanity check

For a strong monoprotic acid, tenfold dilution increases pH by about 1. For a weak acid, tenfold dilution usually increases pH by about 0.5, provided the weak-acid approximation still applies.

Practical link: measuring Ka

You can determine $K_a$ experimentally by preparing a weak acid solution of known concentration and measuring its pH using a calibrated pH meter.

Good practical technique includes calibrating the pH meter with buffer solutions, rinsing the probe with distilled water between measurements, keeping temperature constant, and recording pH to an appropriate number of decimal places. The measured pH is then converted to $H^+]$ and used in the $K_a$ expression.

Exam technique

In the exam

Always decide first whether the acid is strong or weak; do not automatically set $H^+]$ equal to the acid concentration.
For weak monoprotic acids, start from $Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}$ and use $H^+] = [A^-]$ when appropriate.
Check whether the approximation $\approx c$ is reasonable, especially if $K_a$ is large or the acid is very dilute.

Self review

Check yourself

Why does a weak acid have a higher pH than a strong acid of the same concentration?
How would you calculate pH from $K_a$ and the initial concentration of HA?
What does a larger $K_a$ tell you about the position of equilibrium for HA(aq) ⇌ H⁺(aq) + A⁻(aq)?

What you'll learn

What “weak acid” really means, and how it differs from “dilute”.
How to write and use the acid dissociation constant, $K_a$ .
How to calculate the pH of a weak acid from $K_a$ and concentration.
How to calculate $K_a$ from a measured pH.

Prerequisites: acids, hydrogen ions and pH

The pH scale measures hydrogen ion concentration. Lower pH means higher concentration of H⁺ ions.

Definition

For an aqueous solution,

pH=−log⁡10[H+]\text{pH} = -\log_{10}[H^+]

where $H^+]$ is the hydrogen ion concentration in mol dm⁻³.

You can rearrange this to find hydrogen ion concentration:

[H+]=10−pH[H^+] = 10^{-\text{pH}}

Example

Converting pH to hydrogen ion concentration

A solution has pH 3.40. Find $H^+]$ .

Use the rearranged pH equation:
$[H+]=10−pH[H^+] = 10^{-\text{pH}}$
Substitute pH 3.40:
$H^+] = 10^{-3.40}$
Calculate the concentration:
$dm−3[H^+] = 3.98 \times 10^{-4}\ \text{mol dm}^{-3}$

Strong, weak, concentrated and dilute

The words strong and weak describe how much an acid dissociates into ions in water.

A strong acid fully dissociates in aqueous solution. For example:

HCl(aq) → H⁺(aq) + Cl⁻(aq)

A weak acid only partially dissociates, so an equilibrium is set up. For a general monoprotic acid, HA:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

A monoprotic acid donates one proton per acid molecule. The species A⁻ is the conjugate base, formed after HA has donated H⁺.

Definition

Weak acid

A weak acid is an acid that partially dissociates in aqueous solution, forming an equilibrium mixture of undissociated acid molecules and ions.

At equilibrium, a weak acid solution usually contains mostly HA molecules and relatively few H⁺ and A⁻ ions. This is the key particle-level idea behind weak-acid pH calculations.

Particle diagram showing a weak acid HA partially dissociating into H+ and A- at equilibrium, with the Ka expression

Key Idea

Strength is not concentration

The acid dissociation constant, Ka

For the weak monoprotic acid HA:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

the equilibrium constant is called the acid dissociation constant, written $K_a$ .

Definition

Acid dissociation constant

For a weak acid HA,

Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}

using equilibrium concentrations. A larger $K_a$ means the acid is more dissociated, so it is a stronger weak acid.

For this expression, $K_a$ usually has units of mol dm⁻³ at A-Level, because the numerator has two concentration terms and the denominator has one.

$K_a$ is temperature-dependent, so values should only be compared at the same temperature.

pKa

You may also see pKa, defined as:

pKa=−log⁡10Ka\text{p}K_a = -\log_{10}K_a

A lower pKa means a stronger acid, because it corresponds to a larger $K_a$ .

Example

Writing a Ka expression

Methanoic acid, HCOOH, partially dissociates:

HCOOH(aq) ⇌ H⁺(aq) + HCOO⁻(aq)

Identify the weak acid as HCOOH and the ions produced as H⁺ and HCOO⁻.
Put the product concentrations on the top and the undissociated acid concentration on the bottom:
$Ka=[H+][HCOO−][HCOOH]K_a = \frac{[H^+][\text{HCOO}^-]}{[\text{HCOOH}]}$
Use equilibrium concentrations only; do not use the initial acid concentration unless an approximation has been justified.

Calculating pH of a weak acid from Ka

Suppose you have a weak monoprotic acid HA with initial concentration $c$ .

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

Let the equilibrium concentration of H⁺ be $x$ .

Because the acid is monoprotic:

$H^+] = x$
$A^-] = x$
$[H A] = c - x$

So:

Ka=x2c−xK_a = \frac{x^2}{c - x}

For most weak acid calculations at A-Level, the acid is only slightly dissociated, so $x$ is much smaller than $c$ . That means:

\approx c

So the expression simplifies to:

Ka≈x2cK_a \approx \frac{x^2}{c}

and therefore:

[H+]=x≈Kac[H^+] = x \approx \sqrt{K_a c}

Example

Calculating the pH of a weak acid

Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, CH₃COOH, with $dm−3K_a = 1.74 \times 10^{-5}\ \text{mol dm}^{-3}$ .

Write the dissociation equation and link the concentrations:

CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq)

Since one molecule of acid forms one H⁺ and one CH₃COO⁻, let $H^+] = [CH_3COO^-] = x$ .
Use the weak-acid approximation:
$Ka≈x2cK_a \approx \frac{x^2}{c}$
so:
$\approx \sqrt{K_a c}$
Substitute the values:
$\sqrt{(1.74 \times 10^{-5})(0.100)}$ $\times 10^{-3}\ \text{mol dm}^{-3}$
Convert $H^+]$ to pH:
$pH=−log⁡10(1.32×10−3)\text{pH} = -\log_{10}(1.32 \times 10^{-3})$ $pH=2.88\text{pH} = 2.88$
Check the approximation:
$xc×100=1.32×10−30.100×100=1.32%\frac{x}{c} \times 100 = \frac{1.32 \times 10^{-3}}{0.100} \times 100 = 1.32\%$
This is small, so the approximation is reasonable.

Common Mistake

Treating a weak acid like a strong acid

For a 0.100 mol dm⁻³ weak acid, $H^+]$ is not 0.100 mol dm⁻³. Only a small fraction of the acid molecules dissociate, so you must use $K_a$ to find $H^+]$ .

Calculating Ka from pH

Sometimes you are given the pH of a weak acid solution and asked to find $K_a$ .

The usual method is:

Convert pH to $H^+]$ .
Use the 1:1 ratio to find $A^-]$ .
Work out the remaining $[H A]$ .
Substitute into the $K_a$ expression.

For a weak monoprotic acid with no other source of H⁺:

H^+] = [A^-]

and:

HA] = c - [H^+]

Example

Calculating Ka from pH

A 0.0500 mol dm⁻³ solution of a weak monoprotic acid HA has pH 2.93. Calculate $K_a$ .

Convert pH to hydrogen ion concentration:
$H^+] = 10^{-2.93}$ $dm−3[H^+] = 1.17 \times 10^{-3}\ \text{mol dm}^{-3}$
Use the dissociation ratio:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

Since the ratio is 1:1:
$dm−3[A^-] = 1.17 \times 10^{-3}\ \text{mol dm}^{-3}$
Find the equilibrium concentration of undissociated acid:
$\times 10^{-3}$ $0.0488\ \text{mol dm}^{-3}$
Substitute into the $K_a$ expression:
$Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}$ $Ka=(1.17×10−3)(1.17×10−3)0.0488K_a = \frac{(1.17 \times 10^{-3})(1.17 \times 10^{-3})}{0.0488}$ $dm−3K_a = 2.81 \times 10^{-5}\ \text{mol dm}^{-3}$

The weak-acid approximation

The approximation $\approx c$ works when the acid dissociates only slightly. A common check is whether the calculated dissociation is less than about 5% of the original concentration.

If:

xc×100\frac{x}{c} \times 100

is small, the approximation is acceptable.

Common Mistake

When the approximation breaks

If the calculated $H^+]$ is not small compared with the original acid concentration, do not use $\approx c$ . Use $Ka=x2c−xK_a = \frac{x^2}{c - x}$ and solve the resulting quadratic equation.

Dilution and weak acids

For a weak acid where the approximation is valid:

[H+]≈Kac[H^+] \approx \sqrt{K_a c}

This means $H^+]$ depends on the square root of concentration.

So if you dilute a weak acid by a factor of 10, the hydrogen ion concentration falls by about $10\sqrt{10}$ , not by 10. The pH therefore increases by about 0.50, not 1.00.

Tip

Dilution sanity check

For a strong monoprotic acid, tenfold dilution increases pH by about 1. For a weak acid, tenfold dilution usually increases pH by about 0.5, provided the weak-acid approximation still applies.

Practical link: measuring Ka

You can determine $K_a$ experimentally by preparing a weak acid solution of known concentration and measuring its pH using a calibrated pH meter.

Exam technique

In the exam

Always decide first whether the acid is strong or weak; do not automatically set $H^+]$ equal to the acid concentration.
For weak monoprotic acids, start from $Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}$ and use $H^+] = [A^-]$ when appropriate.
Check whether the approximation $\approx c$ is reasonable, especially if $K_a$ is large or the acid is very dilute.

Self review

Check yourself

Why does a weak acid have a higher pH than a strong acid of the same concentration?
How would you calculate pH from $K_a$ and the initial concentration of HA?
What does a larger $K_a$ tell you about the position of equilibrium for HA(aq) ⇌ H⁺(aq) + A⁻(aq)?

What you'll learn

Prerequisites: acids, hydrogen ions and pH

Strong, weak, concentrated and dilute

The acid dissociation constant, Ka

pKa

Calculating pH of a weak acid from Ka

Calculating Ka from pH

The weak-acid approximation

Dilution and weak acids

Practical link: measuring Ka

Review Ka and weak acids by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Ka and weak acids

What you'll learn

Prerequisites: acids, hydrogen ions and pH

Strong, weak, concentrated and dilute

The acid dissociation constant, Ka

pKa

Calculating pH of a weak acid from Ka

Calculating Ka from pH

The weak-acid approximation

Dilution and weak acids

Practical link: measuring Ka

Review Ka and weak acids by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Ka and weak acids

What you'll learn

Prerequisites: acids, hydrogen ions and pH

Strong, weak, concentrated and dilute

The acid dissociation constant, Ka

pKa

Calculating pH of a weak acid from Ka

Calculating Ka from pH

The weak-acid approximation

Dilution and weak acids

Practical link: measuring Ka

Review Ka and weak acids by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Ka and weak acids Revision Guide

Ka and weak acids

What you'll learn

Prerequisites: acids, hydrogen ions and pH

Strong, weak, concentrated and dilute

The acid dissociation constant, Ka

pKa

Calculating pH of a weak acid from Ka

Calculating Ka from pH

The weak-acid approximation

Dilution and weak acids

Practical link: measuring Ka

Review Ka and weak acids by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Ka and weak acids Revision Guide