# Brønsted-Lowry Theory

> Acids as proton donors and bases as proton acceptors, conjugate pairs, amphiprotic substances, strong and weak versus concentrated and dilute, pH and pOH, Kw, Ka and pKa.

From HSC Chemistry on WhizzNotes, free study notes for NSW students: https://whizznotes.com/chemistry/bronsted-lowry-theory/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: Passing the ball
In a game of netball, the ball never stays still. One player passes it, another catches it, and that player passes it on again.

In 1923, Johannes Brønsted in Denmark and Thomas Lowry in England had the same idea in the same year: an acid-base reaction is just a **pass**. The "ball" is a hydrogen ion, which is a single proton. The acid throws it, the base catches it. This simple picture explains things the older models could not, including why ammonia, which has no hydroxide in it, is a base.

**In short:** In Brønsted-Lowry theory, an acid donates a hydrogen ion ($\mathrm{H^+}$) and a base accepts one, and the two differ by one $\mathrm{H^+}$ as a conjugate pair. The pH scale measures acidity, $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$, and a strong acid ionises completely while a weak acid only partly ionises.

## Proton donors and acceptors

A hydrogen atom that loses its electron is just a **proton**, $\mathrm{H^+}$.

- A **Brønsted-Lowry acid** is a **proton donor**.
- A **Brønsted-Lowry base** is a **proton acceptor**.

In water, a proton never floats around alone. It joins a water molecule to make the **hydronium ion**, $\mathrm{H_3O^+}$:

$$\mathrm{HCl(g)} + \mathrm{H_2O(l)} \rightarrow \mathrm{H_3O^+(aq)} + \mathrm{Cl^-(aq)}$$

HCl donates a proton (acid). Water accepts it (base).

Now ammonia:

$$\mathrm{NH_3(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{NH_4^+(aq)} + \mathrm{OH^-(aq)}$$

Ammonia accepts a proton, so it is a base, even though it contains no $\mathrm{OH^-}$. Here water **donates** the proton, so water is acting as an acid.

The Brønsted-Lowry model also works **without water**, for example in the gas phase: $\mathrm{NH_3(g)} + \mathrm{HCl(g)} \rightarrow \mathrm{NH_4Cl(s)}$.

## Conjugate pairs

When an acid gives away a proton, what is left can take a proton back. It is the acid's **conjugate base**.

$$\underset{\text{acid}}{\mathrm{CH_3COOH}} + \underset{\text{base}}{\mathrm{H_2O}} \rightleftharpoons \underset{\text{conjugate base}}{\mathrm{CH_3COO^-}} + \underset{\text{conjugate acid}}{\mathrm{H_3O^+}}$$

A **conjugate pair** differs by exactly **one $\mathrm{H^+}$**:

- $\mathrm{CH_3COOH}$ / $\mathrm{CH_3COO^-}$
- $\mathrm{H_3O^+}$ / $\mathrm{H_2O}$
- $\mathrm{NH_4^+}$ / $\mathrm{NH_3}$

To find a conjugate **base**, take one $\mathrm{H^+}$ off. To find a conjugate **acid**, add one $\mathrm{H^+}$ on.

## Amphiprotic substances

Some species can **donate or accept** a proton. They are **amphiprotic**.

$\mathrm{HCO_3^-}$ as an acid: $\mathrm{HCO_3^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{CO_3^{2-}} + \mathrm{H_3O^+}$

$\mathrm{HCO_3^-}$ as a base: $\mathrm{HCO_3^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_2CO_3} + \mathrm{OH^-}$

Water, $\mathrm{HSO_4^-}$, $\mathrm{H_2PO_4^-}$ and $\mathrm{HPO_4^{2-}}$ are also amphiprotic. A species that is amphiprotic must have an H to give **and** be able to accept one.

## Strong and weak, concentrated and dilute

These are **different ideas**. Do not mix them up.

- **Strong** acid: ionises **completely** in water ($\mathrm{HCl}$, $\mathrm{HNO_3}$, $\mathrm{H_2SO_4}$ first proton). One-way arrow.
- **Weak** acid: ionises only **partly** ($\mathrm{CH_3COOH}$, $\mathrm{HF}$, $\mathrm{H_2CO_3}$). Equilibrium arrow.
- **Concentrated**: lots of acid per litre.
- **Dilute**: little acid per litre.

So "dilute strong acid" and "concentrated weak acid" both make sense.

The stronger an acid, the **weaker** its conjugate base. $\mathrm{Cl^-}$ (from strong HCl) has no real tendency to take a proton back.

**Monoprotic** acids donate one proton (HCl), **diprotic** two ($\mathrm{H_2SO_4}$), **triprotic** three ($\mathrm{H_3PO_4}$).

## pH and pOH

$$\mathrm{pH} = -\log_{10}[\mathrm{H_3O^+}] \qquad [\mathrm{H_3O^+}] = 10^{-\mathrm{pH}}$$

$$\mathrm{pOH} = -\log_{10}[\mathrm{OH^-}] \qquad \mathrm{pH} + \mathrm{pOH} = 14 \text{ (at 25 °C)}$$

- Each pH unit is a **10 times** change in $[\mathrm{H_3O^+}]$. pH 3 is 10 times more acidic than pH 4.
- Diluting a strong acid 10 times raises the pH by 1.

### Worked example: pH of strong acids and bases
0.010 mol/L HCl: $[\mathrm{H_3O^+}] = 0.010$, so pH = 2.0.

0.0050 mol/L $\mathrm{Ba(OH)_2}$: each unit gives two $\mathrm{OH^-}$, so $[\mathrm{OH^-}] = 0.010$, pOH = 2.0 and pH = 12.0.

## The ionisation of water, Kw

Pure water ionises a tiny bit (water is amphiprotic):

$$\mathrm{H_2O(l)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{OH^-(aq)}$$

$$K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14} \text{ at 25 °C}$$

In pure water, $[\mathrm{H_3O^+}] = [\mathrm{OH^-}] = 1.0 \times 10^{-7}$, so pH = 7. The ionisation is endothermic, so at higher temperatures $K_w$ is larger and **neutral pH is below 7**, though the water is still neutral because $[\mathrm{H_3O^+}] = [\mathrm{OH^-}]$.

## Ka and pKa

For a weak acid HA:

$$\mathrm{HA} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-} \qquad K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$$

$$\mathrm{p}K_a = -\log_{10} K_a$$

- **Larger** $K_a$ (smaller $\mathrm{p}K_a$) = **stronger** acid.
- Water is left out of the expression (pure liquid).

### Worked example: pH of a weak acid
0.100 mol/L acetic acid, $K_a = 1.8 \times 10^{-5}$.

Assume a little ionises, so $[\mathrm{HA}] \approx 0.100$ and $[\mathrm{H_3O^+}] = [\mathrm{A^-}] = x$:

$x^2 = 1.8 \times 10^{-5} \times 0.100 = 1.8 \times 10^{-6}$, so $x = 1.34 \times 10^{-3}$.

pH $= -\log(1.34 \times 10^{-3}) = 2.87$.

Percentage ionised $= \dfrac{1.34 \times 10^{-3}}{0.100} \times 100 = 1.3\%$.

**Mind map**

```text
Brønsted-Lowry
  Definitions
    Acid: proton donor
    Base: proton acceptor
    H3O+ hydronium
  Conjugate pairs
    Differ by one H+
    Strong acid, weak conjugate base
  Amphiprotic
    HCO3−, H2O, HSO4−
  Strength
    Strong: fully ionised
    Weak: partly ionised
    Not the same as concentration
  pH
    pH = −log[H3O+]
    pH + pOH = 14
    Kw = 1.0 × 10⁻¹⁴
  Ka
    Bigger Ka, stronger
    pKa = −log Ka
```

### Year 11: Logs on your calculator
$\mathrm{pH} = -\log[\mathrm{H^+}]$: press **(−) log** then the concentration. For $[\mathrm{H^+}] = 3.5 \times 10^{-4}$, pH $= 3.46$.

Going back: $[\mathrm{H^+}] = 10^{-\mathrm{pH}}$. Press **SHIFT log** (which is $10^x$), then **(−)** and the pH. For pH 4.72, $[\mathrm{H^+}] = 1.9 \times 10^{-5}$ mol/L.

**Significant figures.** The number of decimal places in a pH equals the number of significant figures in the concentration.

### Year 12: Bases, salts and Kb
For a weak base B: $\mathrm{B} + \mathrm{H_2O} \rightleftharpoons \mathrm{BH^+} + \mathrm{OH^-}$, with $K_b = \dfrac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]}$.

For any conjugate pair: $K_a \times K_b = K_w$, so $\mathrm{p}K_a + \mathrm{p}K_b = 14$.

**Why salts are acidic or basic.** In sodium acetate, the acetate ion is a weak base:

$$\mathrm{CH_3COO^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH_3COOH} + \mathrm{OH^-}$$

$K_b = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$. For 0.10 mol/L, $[\mathrm{OH^-}] = \sqrt{5.6 \times 10^{-11}} = 7.5 \times 10^{-6}$, so pOH = 5.13 and **pH = 8.87**.

In ammonium chloride, $\mathrm{NH_4^+} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_3} + \mathrm{H_3O^+}$, so the solution is acidic. $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ do not react with water, so NaCl is neutral.

**Finding Ka from pH.** A 0.10 mol/L weak acid has pH 3.00. $[\mathrm{H^+}] = 1.0 \times 10^{-3}$, so $K_a = \dfrac{(1.0 \times 10^{-3})^2}{0.10 - 0.001} = 1.0 \times 10^{-5}$.

**pH of a weak base.** Use $K_b$ the same way as $K_a$: for 0.250 mol/L methanamine ($K_b = 4.38 \times 10^{-4}$), $[\mathrm{OH^-}] = \sqrt{4.38 \times 10^{-4} \times 0.250} = 0.0105$, so pOH = 1.98 and pH = 12.02. Its salt, methylammonium chloride, is **acidic**, so it could never have pH 12.

**Percentage ionised from pH.** A 0.050 mol/L acid at pH 3.035 has $[\mathrm{H^+}] = 9.23 \times 10^{-4}$, so $\dfrac{9.23 \times 10^{-4}}{0.050} \times 100 = 1.8\%$ is ionised. Diluting a weak acid makes a **larger** percentage ionise, but the pH still rises.

**Mixing a strong acid and a strong base.** Find the moles of $\mathrm{H^+}$ and $\mathrm{OH^-}$ (remember $\mathrm{Ba(OH)_2}$ gives two $\mathrm{OH^-}$ and $\mathrm{H_2SO_4}$ two $\mathrm{H^+}$), subtract, then divide what is left over by the **total** volume. 0.0030 mol $\mathrm{H^+}$ and 0.0020 mol $\mathrm{OH^-}$ in 100 mL leave 0.0010 mol in 0.100 L: $[\mathrm{H^+}] = 0.010$, pH = 2.00.

**Raising the pH of a strong base.** To go from pH 12.0 to 13.0 in 1 L of KOH, $[\mathrm{OH^-}]$ must go from 0.010 to 0.10 mol/L. Add 0.090 mol, which is 5.05 g of KOH.

**Neutral is not always pH 7.** At 60 °C, $K_w = 9.6 \times 10^{-14}$, so $[\mathrm{H_3O^+}] = \sqrt{9.6 \times 10^{-14}} = 3.1 \times 10^{-7}$ and pH = 6.51. The water is still neutral.

## Quick summary

- Brønsted-Lowry: acid donates $\mathrm{H^+}$, base accepts $\mathrm{H^+}$. Works without water and explains ammonia.
- Conjugate pairs differ by one $\mathrm{H^+}$. Amphiprotic species can do both.
- Strong/weak is about how much ionises. Concentrated/dilute is about how much is dissolved.
- $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$, $\mathrm{pH} + \mathrm{pOH} = 14$, $K_w = 1.0 \times 10^{-14}$ at 25 °C.
- $K_a$ measures acid strength: larger $K_a$, smaller $\mathrm{p}K_a$, stronger acid.
- Weak acid: $[\mathrm{H^+}] = \sqrt{K_a \times c}$.

## Quiz

Try these questions from the chapter quiz.

1. A Brønsted-Lowry acid is a:
   - A. Proton donor
   - B. Proton acceptor
   - C. Hydroxide donor
   - D. Electron donor
   - Answer: A. Proton donor. Acids give away $\mathrm{H^+}$.

2. In $\mathrm{NH_3} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_4^+} + \mathrm{OH^-}$, water acts as:
   - A. An acid
   - B. A base
   - C. A salt
   - D. A catalyst
   - Answer: A. An acid. Water gives a proton to ammonia.

3. The conjugate base of $\mathrm{HSO_4^-}$ is:
   - A. $\mathrm{H_2SO_4}$
   - B. $\mathrm{SO_4^{2-}}$
   - C. $\mathrm{HSO_4}$
   - D. $\mathrm{OH^-}$
   - Answer: B. $\mathrm{SO_4^{2-}}$. Take one $\mathrm{H^+}$ off.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/chemistry/bronsted-lowry-theory/#quiz