Acid-base equilibria

What you'll learn

How Lowry–Brønsted theory explains acids, bases and conjugate pairs.
How to use $K_a$ , $pK_a$ , $K_w$ and pH in quantitative calculations.
How titration curves, indicators, buffers and salt hydrolysis are linked.
How a pH-probe titration is carried out and evaluated.

1. The starting point: proton transfer

Acid–base equilibria are all about the movement of protons, meaning hydrogen ions, $H+\mathrm{H^+}$ . In water, $H+\mathrm{H^+}$ is usually attached to a water molecule as $H3O+\mathrm{H_3O^+}$ , but A-Level calculations often write simply $H+\mathrm{H^+}$ .

Definition

Lowry–Brønsted acids and bases

A Lowry–Brønsted acid is a proton donor. A Lowry–Brønsted base is a proton acceptor.

For example:

$NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}$

Ammonia accepts a proton, so it acts as a base. Water donates a proton, so it acts as an acid.

A conjugate acid–base pair contains two species that differ by one proton.

Example

Identifying conjugate pairs

For the equilibrium $NH3+H2O⇌NH4++OH−\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}$ :

Track the proton transfer: water loses $H+\mathrm{H^+}$ to form $OH−\mathrm{OH^-}$ , while ammonia gains $H+\mathrm{H^+}$ to form $NH4+\mathrm{NH_4^+}$ .
The acid is $H2O\mathrm{H_2O}$ and its conjugate base is $OH−\mathrm{OH^-}$ .
The base is $NH3\mathrm{NH_3}$ and its conjugate acid is $NH4+\mathrm{NH_4^+}$ .

2. Strong and weak are not the same as concentrated and dilute

A strong acid fully dissociates in water. A weak acid only partially dissociates, setting up an equilibrium.

Examples:

Strong acid: $HCl(aq)→H+(aq)+Cl−(aq)\mathrm{HCl(aq) \to H^+(aq) + Cl^-(aq)}$
Weak acid: $CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$

A strong base fully releases hydroxide ions in water, such as sodium hydroxide:

$NaOH(aq)→Na+(aq)+OH−(aq)\mathrm{NaOH(aq) \to Na^+(aq) + OH^-(aq)}$

A weak base only partially reacts with water, such as ammonia:

$NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}$

Key Idea

Strength is about dissociation

Strong/weak describes the extent of ionisation or proton transfer. Concentrated/dilute describes how much solute is dissolved per dm³ of solution.

Common Mistake

Calling dilute acids weak

A dilute hydrochloric acid solution is still a strong acid because each HCl molecule dissociates fully. It is dilute because the concentration is low.

3. pH and strong acid calculations

pH measures hydrogen ion concentration:

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

You can reverse this using:

[H+]=10−pH[\mathrm{H^+}] = 10^{-\text{pH}}

For a strong monoprotic acid such as HCl or HNO₃, the acid fully dissociates, so the hydrogen ion concentration equals the acid concentration.

Example

Finding the pH of a strong acid

Find the pH of 0.0250 mol dm⁻³ hydrochloric acid.

HCl is a strong monoprotic acid, so one mole of HCl gives one mole of $H+\mathrm{H^+}$ . Therefore $dm−3[\mathrm{H^+}] = 0.0250\ \mathrm{mol\ dm^{-3}}$ .
Substitute into the pH equation:
$pH=−log⁡(0.0250)=1.60\text{pH} = -\log(0.0250) = 1.60$
Check the size: 0.0100 mol dm⁻³ acid has pH 2.00, so a more concentrated acid having pH 1.60 is sensible.

4. The acid dissociation constant, $K_a$

For a weak acid, HA:

$HA(aq)⇌H+(aq)+A−(aq)\mathrm{HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)}$

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

Definition

Acid dissociation constant

For $HA⇌H++A−\mathrm{HA \rightleftharpoons H^+ + A^-}$ :

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

For this simple equilibrium, $K_a$ has units of mol dm⁻³ at A-Level.

A larger $K_a$ means the acid dissociates more, so it is a stronger acid. Chemists often use $pK_a$ :

pK_a = -\log K_a

So:

lower $pK_a$ means stronger acid
higher $pK_a$ means weaker acid

For many weak acid calculations, if the initial acid concentration is $c$ , then:

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

This works when the acid is only slightly dissociated.

Example

Calculating the pH of a weak acid

Ethanoic acid has $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\ dm^{-3}}$ . Calculate the pH of 0.100 mol dm⁻³ ethanoic acid.

Write the weak acid approximation:
$[H+]≈Kac[\mathrm{H^+}] \approx \sqrt{K_a c}$
Substitute the values:
$dm−3[\mathrm{H^+}] = \sqrt{(1.74 \times 10^{-5})(0.100)} = 1.32 \times 10^{-3}\ \mathrm{mol\ dm^{-3}}$
Calculate pH:
$pH=−log⁡(1.32×10−3)=2.88\text{pH} = -\log(1.32 \times 10^{-3}) = 2.88$
Check the approximation: the fraction dissociated is about 1.32%, so treating the undissociated acid concentration as approximately 0.100 mol dm⁻³ is reasonable.

Common Mistake

When the weak acid shortcut can fail

The approximation $[H+]≈Kac[\mathrm{H^+}] \approx \sqrt{K_a c}$ becomes less reliable if the acid is very dilute or if the acid is not very weak. In those cases, you may need to solve the full quadratic expression.

5. The ionic product of water, $K_w$

Water very slightly ionises:

$H2O(l)⇌H+(aq)+OH−(aq)\mathrm{H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)}$

The equilibrium expression is:

Kw=[H+][OH−]K_w = [\mathrm{H^+}][\mathrm{OH^-}]

At 298 K:

dm−6K_w = 1.00 \times 10^{-14}\ \mathrm{mol^2\ dm^{-6}}

In pure water at 298 K, $[H+]=[OH−][\mathrm{H^+}] = [\mathrm{OH^-}]$ , so pH is 7.00.

Tip

Neutral does not always mean pH 7

$K_w$ changes with temperature. Neutral means $[H+]=[OH−][\mathrm{H^+}] = [\mathrm{OH^-}]$ , but the pH of neutrality is only exactly 7.00 at 298 K.

Strong base calculations use $K_w$ because a strong base directly gives $OH−\mathrm{OH^-}$ , then $K_w$ gives $H+\mathrm{H^+}$ .

Example

Finding the pH of a strong base

Calculate the pH of 0.0200 mol dm⁻³ NaOH at 298 K.

NaOH is a strong base, so $dm−3[\mathrm{OH^-}] = 0.0200\ \mathrm{mol\ dm^{-3}}$ .
Use $Kw=[H+][OH−]K_w = [\mathrm{H^+}][\mathrm{OH^-}]$ :
$dm−3[\mathrm{H^+}] = \frac{1.00 \times 10^{-14}}{0.0200} = 5.00 \times 10^{-13}\ \mathrm{mol\ dm^{-3}}$
Calculate pH:
$pH=−log⁡(5.00×10−13)=12.30\text{pH} = -\log(5.00 \times 10^{-13}) = 12.30$

6. Titration curves and equivalence points

A titration curve shows how pH changes as one solution is added to another. The equivalence point is where the acid and base have reacted in the exact stoichiometric ratio. It is not always pH 7.

The curves below show base being added to acid.

Four acid-base titration curves showing strong/weak acid and base combinations, equivalence points, buffer region and indicator ranges

Key shapes:

Strong acid + strong base: steep vertical section centred around pH 7.
Weak acid + strong base: starts at a higher pH; has a buffer region; equivalence point is above pH 7 because the conjugate base hydrolyses.
Strong acid + weak base: equivalence point is below pH 7 because the conjugate acid hydrolyses.
Weak acid + weak base: no sharp vertical section, so visual indicators are unreliable.

Practical: titration using a pH probe

A typical method:

Calibrate the pH probe using standard buffer solutions, usually pH 4, 7 and/or 10.
Use a pipette to place a known volume of acid or base into a conical flask.
Add the other solution from a burette in measured portions, stirring after each addition.
Record the pH once the reading is stable.
Add smaller volumes near the rapid pH change.
Plot pH against volume added and find the equivalence point from the steepest part of the curve.

Main sources of error include poor probe calibration, not waiting for a stable pH reading, temperature changes affecting $K_w$ , and overshooting near the equivalence point.

7. Choosing indicators

An acid–base indicator is a weak acid or weak base with different colours in its acid and base forms. It changes colour over a narrow pH range.

A suitable indicator must change colour within the steep vertical section of the titration curve.

Example

Choosing an indicator

Choose an indicator for titrating a weak acid with a strong base.

A weak acid–strong base titration has an equivalence point above pH 7.
Phenolphthalein changes around pH 8.3–10.0, which usually lies within the steep section for this titration.
Methyl orange changes around pH 3.1–4.4, which is too early, before the equivalence point.
Therefore phenolphthalein is suitable; for weak acid–weak base titrations, a pH probe is preferred because there is no sharp vertical section.

8. Buffer solutions

A buffer solution resists changes in pH when small amounts of acid or alkali are added.

An acidic buffer contains:

a weak acid, HA
its conjugate base, A⁻, often from a soluble salt such as sodium ethanoate

The buffer works because both acid and base components are present:

Added acid is removed by $A−+H+→HA\mathrm{A^- + H^+ \to HA}$ .
Added alkali is removed by $HA+OH−→A−+H2O\mathrm{HA + OH^- \to A^- + H_2O}$ .

For an acidic buffer:

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

Rearranging:

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

Or in pH form:

pH=pKa+log⁡([A−][HA])\text{pH} = pK_a + \log\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)

Tip

Use moles in buffer mixtures

If HA and A⁻ are in the same final solution volume, the volume cancels in the ratio, so you can use moles instead of concentrations.

Example

Calculating the pH of a buffer after neutralisation

25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid is mixed with 10.0 cm³ of 0.100 mol dm⁻³ NaOH. For ethanoic acid, $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\ dm^{-3}}$ . Calculate the buffer pH.

Calculate initial amounts:
$moln(\mathrm{CH_3COOH}) = 0.100 \times \frac{25.0}{1000} = 2.50 \times 10^{-3}\ \mathrm{mol}$ $moln(\mathrm{OH^-}) = 0.100 \times \frac{10.0}{1000} = 1.00 \times 10^{-3}\ \mathrm{mol}$
Neutralisation converts acid into conjugate base:

$CH3COOH+OH−→CH3COO−+H2O\mathrm{CH_3COOH + OH^- \to CH_3COO^- + H_2O}$

So remaining HA is $1.50 \times 10^{-3}\ \mathrm{mol}$ and formed A⁻ is $1.00 \times 10^{-3}\ \mathrm{mol}$ .
Calculate $pK_a$ :
$pKa=−log⁡(1.74×10−5)=4.76pK_a = -\log(1.74 \times 10^{-5}) = 4.76$
Substitute into the buffer equation using the mole ratio:
$pH=4.76+log⁡(1.00×10−31.50×10−3)=4.58\text{pH} = 4.76 + \log\left(\frac{1.00 \times 10^{-3}}{1.50 \times 10^{-3}}\right) = 4.58$

Buffers are vital in living systems because enzyme activity is pH dependent. Blood is buffered mainly by the carbonic acid/hydrogencarbonate system, helping maintain pH close to 7.4. Industrially, buffers are used in fermentation, medicines, food production, dyeing and analytical calibration.

9. Hydrolysis of salts

Salt hydrolysis means ions from a salt react with water to produce $H+\mathrm{H^+}$ or $OH−\mathrm{OH^-}$ , changing the pH.

A salt from a strong acid and strong base, such as NaCl, gives a roughly neutral solution. The ions do not hydrolyse significantly.
A salt from a strong acid and weak base, such as NH₄Cl, gives an acidic solution:

$NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)\mathrm{NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)}$
A salt from a weak acid and strong base, such as sodium ethanoate, gives an alkaline solution:

$CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)\mathrm{CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)}$

Example

Predicting salt solution pH

Predict whether sodium ethanoate solution is acidic, neutral or alkaline.

Identify the parent acid and base: sodium ethanoate comes from ethanoic acid, a weak acid, and sodium hydroxide, a strong base.
Sodium ions do not hydrolyse significantly, but ethanoate ions can accept protons from water.
Hydrolysis forms $OH−\mathrm{OH^-}$ , so the solution is alkaline.

Exam technique

In the exam

Decide first whether the acid/base is strong or weak; this tells you whether to use stoichiometry, $K_a$ , $K_w$ or a buffer equation.
For weak acid and buffer calculations, keep track of equilibrium concentrations or mole ratios carefully before substituting into equations.
In titration questions, use the curve shape and equivalence-point pH to choose the indicator, not just the names of the reactants.
Always check whether pH 7 is actually justified; equivalence point and neutrality are different ideas.

Self review

Check yourself

Why does a weak acid–strong base titration have an equivalence point above pH 7?
How would you calculate the pH of a strong base using $K_w$ ?
What two components must an acidic buffer contain, and how does each component respond to added acid or alkali?

What you'll learn

How Lowry–Brønsted theory explains acids, bases and conjugate pairs.
How to use $K_a$ , $pK_a$ , $K_w$ and pH in quantitative calculations.
How titration curves, indicators, buffers and salt hydrolysis are linked.
How a pH-probe titration is carried out and evaluated.

1. The starting point: proton transfer

Definition

Lowry–Brønsted acids and bases

A Lowry–Brønsted acid is a proton donor. A Lowry–Brønsted base is a proton acceptor.

For example:

$NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}$

Ammonia accepts a proton, so it acts as a base. Water donates a proton, so it acts as an acid.

A conjugate acid–base pair contains two species that differ by one proton.

Example

Identifying conjugate pairs

For the equilibrium $NH3+H2O⇌NH4++OH−\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}$ :

Track the proton transfer: water loses $H+\mathrm{H^+}$ to form $OH−\mathrm{OH^-}$ , while ammonia gains $H+\mathrm{H^+}$ to form $NH4+\mathrm{NH_4^+}$ .
The acid is $H2O\mathrm{H_2O}$ and its conjugate base is $OH−\mathrm{OH^-}$ .
The base is $NH3\mathrm{NH_3}$ and its conjugate acid is $NH4+\mathrm{NH_4^+}$ .

2. Strong and weak are not the same as concentrated and dilute

A strong acid fully dissociates in water. A weak acid only partially dissociates, setting up an equilibrium.

Examples:

Strong acid: $HCl(aq)→H+(aq)+Cl−(aq)\mathrm{HCl(aq) \to H^+(aq) + Cl^-(aq)}$
Weak acid: $CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$

A strong base fully releases hydroxide ions in water, such as sodium hydroxide:

$NaOH(aq)→Na+(aq)+OH−(aq)\mathrm{NaOH(aq) \to Na^+(aq) + OH^-(aq)}$

A weak base only partially reacts with water, such as ammonia:

$NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}$

Key Idea

Strength is about dissociation

Strong/weak describes the extent of ionisation or proton transfer. Concentrated/dilute describes how much solute is dissolved per dm³ of solution.

Common Mistake

Calling dilute acids weak

A dilute hydrochloric acid solution is still a strong acid because each HCl molecule dissociates fully. It is dilute because the concentration is low.

3. pH and strong acid calculations

pH measures hydrogen ion concentration:

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

You can reverse this using:

[H+]=10−pH[\mathrm{H^+}] = 10^{-\text{pH}}

For a strong monoprotic acid such as HCl or HNO₃, the acid fully dissociates, so the hydrogen ion concentration equals the acid concentration.

Example

Finding the pH of a strong acid

Find the pH of 0.0250 mol dm⁻³ hydrochloric acid.

HCl is a strong monoprotic acid, so one mole of HCl gives one mole of $H+\mathrm{H^+}$ . Therefore $dm−3[\mathrm{H^+}] = 0.0250\ \mathrm{mol\ dm^{-3}}$ .
Substitute into the pH equation:
$pH=−log⁡(0.0250)=1.60\text{pH} = -\log(0.0250) = 1.60$
Check the size: 0.0100 mol dm⁻³ acid has pH 2.00, so a more concentrated acid having pH 1.60 is sensible.

4. The acid dissociation constant, $K_a$

For a weak acid, HA:

$HA(aq)⇌H+(aq)+A−(aq)\mathrm{HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)}$

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

Definition

Acid dissociation constant

For $HA⇌H++A−\mathrm{HA \rightleftharpoons H^+ + A^-}$ :

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

For this simple equilibrium, $K_a$ has units of mol dm⁻³ at A-Level.

A larger $K_a$ means the acid dissociates more, so it is a stronger acid. Chemists often use $pK_a$ :

pK_a = -\log K_a

So:

lower $pK_a$ means stronger acid
higher $pK_a$ means weaker acid

For many weak acid calculations, if the initial acid concentration is $c$ , then:

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

This works when the acid is only slightly dissociated.

Example

Calculating the pH of a weak acid

Ethanoic acid has $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\ dm^{-3}}$ . Calculate the pH of 0.100 mol dm⁻³ ethanoic acid.

Write the weak acid approximation:
$[H+]≈Kac[\mathrm{H^+}] \approx \sqrt{K_a c}$
Substitute the values:
$dm−3[\mathrm{H^+}] = \sqrt{(1.74 \times 10^{-5})(0.100)} = 1.32 \times 10^{-3}\ \mathrm{mol\ dm^{-3}}$
Calculate pH:
$pH=−log⁡(1.32×10−3)=2.88\text{pH} = -\log(1.32 \times 10^{-3}) = 2.88$
Check the approximation: the fraction dissociated is about 1.32%, so treating the undissociated acid concentration as approximately 0.100 mol dm⁻³ is reasonable.

Common Mistake

When the weak acid shortcut can fail

5. The ionic product of water, $K_w$

Water very slightly ionises:

$H2O(l)⇌H+(aq)+OH−(aq)\mathrm{H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)}$

The equilibrium expression is:

Kw=[H+][OH−]K_w = [\mathrm{H^+}][\mathrm{OH^-}]

At 298 K:

dm−6K_w = 1.00 \times 10^{-14}\ \mathrm{mol^2\ dm^{-6}}

In pure water at 298 K, $[H+]=[OH−][\mathrm{H^+}] = [\mathrm{OH^-}]$ , so pH is 7.00.

Tip

Neutral does not always mean pH 7

$K_w$ changes with temperature. Neutral means $[H+]=[OH−][\mathrm{H^+}] = [\mathrm{OH^-}]$ , but the pH of neutrality is only exactly 7.00 at 298 K.

Strong base calculations use $K_w$ because a strong base directly gives $OH−\mathrm{OH^-}$ , then $K_w$ gives $H+\mathrm{H^+}$ .

Example

Finding the pH of a strong base

Calculate the pH of 0.0200 mol dm⁻³ NaOH at 298 K.

NaOH is a strong base, so $dm−3[\mathrm{OH^-}] = 0.0200\ \mathrm{mol\ dm^{-3}}$ .
Use $Kw=[H+][OH−]K_w = [\mathrm{H^+}][\mathrm{OH^-}]$ :
$dm−3[\mathrm{H^+}] = \frac{1.00 \times 10^{-14}}{0.0200} = 5.00 \times 10^{-13}\ \mathrm{mol\ dm^{-3}}$
Calculate pH:
$pH=−log⁡(5.00×10−13)=12.30\text{pH} = -\log(5.00 \times 10^{-13}) = 12.30$

6. Titration curves and equivalence points

The curves below show base being added to acid.

Four acid-base titration curves showing strong/weak acid and base combinations, equivalence points, buffer region and indicator ranges

Key shapes:

Strong acid + strong base: steep vertical section centred around pH 7.
Weak acid + strong base: starts at a higher pH; has a buffer region; equivalence point is above pH 7 because the conjugate base hydrolyses.
Strong acid + weak base: equivalence point is below pH 7 because the conjugate acid hydrolyses.
Weak acid + weak base: no sharp vertical section, so visual indicators are unreliable.

Practical: titration using a pH probe

A typical method:

Calibrate the pH probe using standard buffer solutions, usually pH 4, 7 and/or 10.
Use a pipette to place a known volume of acid or base into a conical flask.
Add the other solution from a burette in measured portions, stirring after each addition.
Record the pH once the reading is stable.
Add smaller volumes near the rapid pH change.
Plot pH against volume added and find the equivalence point from the steepest part of the curve.

Main sources of error include poor probe calibration, not waiting for a stable pH reading, temperature changes affecting $K_w$ , and overshooting near the equivalence point.

7. Choosing indicators

An acid–base indicator is a weak acid or weak base with different colours in its acid and base forms. It changes colour over a narrow pH range.

A suitable indicator must change colour within the steep vertical section of the titration curve.

Example

Choosing an indicator

Choose an indicator for titrating a weak acid with a strong base.

A weak acid–strong base titration has an equivalence point above pH 7.
Phenolphthalein changes around pH 8.3–10.0, which usually lies within the steep section for this titration.
Methyl orange changes around pH 3.1–4.4, which is too early, before the equivalence point.
Therefore phenolphthalein is suitable; for weak acid–weak base titrations, a pH probe is preferred because there is no sharp vertical section.

8. Buffer solutions

A buffer solution resists changes in pH when small amounts of acid or alkali are added.

An acidic buffer contains:

a weak acid, HA
its conjugate base, A⁻, often from a soluble salt such as sodium ethanoate

The buffer works because both acid and base components are present:

Added acid is removed by $A−+H+→HA\mathrm{A^- + H^+ \to HA}$ .
Added alkali is removed by $HA+OH−→A−+H2O\mathrm{HA + OH^- \to A^- + H_2O}$ .

For an acidic buffer:

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

Rearranging:

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

Or in pH form:

pH=pKa+log⁡([A−][HA])\text{pH} = pK_a + \log\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)

Tip

Use moles in buffer mixtures

If HA and A⁻ are in the same final solution volume, the volume cancels in the ratio, so you can use moles instead of concentrations.

Example

Calculating the pH of a buffer after neutralisation

25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid is mixed with 10.0 cm³ of 0.100 mol dm⁻³ NaOH. For ethanoic acid, $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\ dm^{-3}}$ . Calculate the buffer pH.

Calculate initial amounts:
$moln(\mathrm{CH_3COOH}) = 0.100 \times \frac{25.0}{1000} = 2.50 \times 10^{-3}\ \mathrm{mol}$ $moln(\mathrm{OH^-}) = 0.100 \times \frac{10.0}{1000} = 1.00 \times 10^{-3}\ \mathrm{mol}$
Neutralisation converts acid into conjugate base:

$CH3COOH+OH−→CH3COO−+H2O\mathrm{CH_3COOH + OH^- \to CH_3COO^- + H_2O}$

So remaining HA is $1.50 \times 10^{-3}\ \mathrm{mol}$ and formed A⁻ is $1.00 \times 10^{-3}\ \mathrm{mol}$ .
Calculate $pK_a$ :
$pKa=−log⁡(1.74×10−5)=4.76pK_a = -\log(1.74 \times 10^{-5}) = 4.76$
Substitute into the buffer equation using the mole ratio:
$pH=4.76+log⁡(1.00×10−31.50×10−3)=4.58\text{pH} = 4.76 + \log\left(\frac{1.00 \times 10^{-3}}{1.50 \times 10^{-3}}\right) = 4.58$

9. Hydrolysis of salts

Salt hydrolysis means ions from a salt react with water to produce $H+\mathrm{H^+}$ or $OH−\mathrm{OH^-}$ , changing the pH.

A salt from a strong acid and strong base, such as NaCl, gives a roughly neutral solution. The ions do not hydrolyse significantly.
A salt from a strong acid and weak base, such as NH₄Cl, gives an acidic solution:

$NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)\mathrm{NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)}$
A salt from a weak acid and strong base, such as sodium ethanoate, gives an alkaline solution:

$CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)\mathrm{CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)}$

Example

Predicting salt solution pH

Predict whether sodium ethanoate solution is acidic, neutral or alkaline.

Identify the parent acid and base: sodium ethanoate comes from ethanoic acid, a weak acid, and sodium hydroxide, a strong base.
Sodium ions do not hydrolyse significantly, but ethanoate ions can accept protons from water.
Hydrolysis forms $OH−\mathrm{OH^-}$ , so the solution is alkaline.

Exam technique

In the exam

Decide first whether the acid/base is strong or weak; this tells you whether to use stoichiometry, $K_a$ , $K_w$ or a buffer equation.
For weak acid and buffer calculations, keep track of equilibrium concentrations or mole ratios carefully before substituting into equations.
In titration questions, use the curve shape and equivalence-point pH to choose the indicator, not just the names of the reactants.
Always check whether pH 7 is actually justified; equivalence point and neutrality are different ideas.

Self review

Check yourself

Why does a weak acid–strong base titration have an equivalence point above pH 7?
How would you calculate the pH of a strong base using $K_w$ ?
What two components must an acidic buffer contain, and how does each component respond to added acid or alkali?

What you'll learn

1. The starting point: proton transfer

2. Strong and weak are not the same as concentrated and dilute

3. pH and strong acid calculations

4. The acid dissociation constant, $K_a$

5. The ionic product of water, $K_w$

6. Titration curves and equivalence points

Practical: titration using a pH probe

7. Choosing indicators

8. Buffer solutions

9. Hydrolysis of salts

Review Acid-base equilibria by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Acid-base equilibria

What you'll learn

1. The starting point: proton transfer

2. Strong and weak are not the same as concentrated and dilute

3. pH and strong acid calculations

4. The acid dissociation constant, $K_a$

5. The ionic product of water, $K_w$

6. Titration curves and equivalence points

Practical: titration using a pH probe

7. Choosing indicators

8. Buffer solutions

9. Hydrolysis of salts

Review Acid-base equilibria by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Acid-base equilibria

What you'll learn

1. The starting point: proton transfer

2. Strong and weak are not the same as concentrated and dilute

3. pH and strong acid calculations

4. The acid dissociation constant, KaK_aKa​

5. The ionic product of water, KwK_wKw​

6. Titration curves and equivalence points

Practical: titration using a pH probe

7. Choosing indicators

8. Buffer solutions

9. Hydrolysis of salts

Review Acid-base equilibria by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Acid-base equilibria Revision Guide

Acid-base equilibria

What you'll learn

1. The starting point: proton transfer

2. Strong and weak are not the same as concentrated and dilute

3. pH and strong acid calculations

4. The acid dissociation constant, KaK_aKa​

5. The ionic product of water, KwK_wKw​

6. Titration curves and equivalence points

Practical: titration using a pH probe

7. Choosing indicators

8. Buffer solutions

9. Hydrolysis of salts

Review Acid-base equilibria by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Acid-base equilibria Revision Guide

4. The acid dissociation constant, $K_a$

5. The ionic product of water, $K_w$

4. The acid dissociation constant, $K_a$

5. The ionic product of water, $K_w$