Brønsted-Lowry Theory

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, $\mathrm{pH}=-\log [{\mathrm{H}}_3{\mathrm{O}}^{+}]$, 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, ${\mathrm{H}}_3{\mathrm{O}}^{+}$:

$$\mathrm{HCl}(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to {\mathrm{H}}_3{\mathrm{O}}^{+}(\mathrm{aq})+\mathrm{C}{\mathrm{l}}^{-}(\mathrm{aq})$$

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

Now ammonia:

$$\mathrm{N}{\mathrm{H}}_3(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\rightleftharpoons \mathrm{N}{\mathrm{H}}_4^{+}(\mathrm{aq})+\mathrm{O}{\mathrm{H}}^{-}(\mathrm{aq})$$

Ammonia accepts a proton, so it is a base, even though it contains no $\mathrm{O}{\mathrm{H}}^{-}$. 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: $\mathrm{N}{\mathrm{H}}_3(\mathrm{g})+\mathrm{HCl}(\mathrm{g})\to \mathrm{N}{\mathrm{H}}_4\mathrm{Cl}(\mathrm{s})$.

Conjugate pairs

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

$$\underset{\text{acid}}{\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}}+\underset{\text{base}}{{\mathrm{H}}_2\mathrm{O}}\rightleftharpoons \underset{\text{conjugate base}}{\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}}+\underset{\text{conjugate acid}}{{\mathrm{H}}_3{\mathrm{O}}^{+}}$$

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

• $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ / $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}$
• ${\mathrm{H}}_3{\mathrm{O}}^{+}$ / ${\mathrm{H}}_2\mathrm{O}$
• $\mathrm{N}{\mathrm{H}}_4^{+}$ / $\mathrm{N}{\mathrm{H}}_3$

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.

$\mathrm{HC}{\mathrm{O}}_3^{-}$ as an acid: $\mathrm{HC}{\mathrm{O}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{C}{\mathrm{O}}_3^{2-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$

$\mathrm{HC}{\mathrm{O}}_3^{-}$ as a base: $\mathrm{HC}{\mathrm{O}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_3+\mathrm{O}{\mathrm{H}}^{-}$

Water, $\mathrm{HS}{\mathrm{O}}_4^{-}$, ${\mathrm{H}}_2\mathrm{P}{\mathrm{O}}_4^{-}$ and $\mathrm{HP}{\mathrm{O}}_4^{2-}$ 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 ($\mathrm{HCl}$, $\mathrm{HN}{\mathrm{O}}_3$, ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ first proton). One-way arrow.
• Weak acid: ionises only partly ($\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$, $\mathrm{HF}$, ${\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_3$). 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. $\mathrm{C}{\mathrm{l}}^{-}$ (from strong HCl) has no real tendency to take a proton back.

Monoprotic acids donate one proton (HCl), diprotic two (${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$), triprotic three (${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$).

pH and pOH

$$\mathrm{pH}=-{\log }_{10}[{\mathrm{H}}_3{\mathrm{O}}^{+}][{\mathrm{H}}_3{\mathrm{O}}^{+}]=10^{-\mathrm{pH}}$$

$$\mathrm{pOH}=-{\log }_{10}[\mathrm{O}{\mathrm{H}}^{-}]\mathrm{pH}+\mathrm{pOH}=14\text{ (at 25 °C)}$$

• Each pH unit is a 10 times change in $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$. pH 3 is 10 times more acidic than pH 4.
• Diluting a strong acid 10 times raises the pH by 1.

The ionisation of water, Kw

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

$${\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}(\mathrm{aq})+\mathrm{O}{\mathrm{H}}^{-}(\mathrm{aq})$$

$$K_w=[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]=1.0\times 10^{-14}\text{ at 25 °C}$$

In pure water, $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=[\mathrm{O}{\mathrm{H}}^{-}]=1.0\times 10^{-7}$, 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 $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=[\mathrm{O}{\mathrm{H}}^{-}]$.

Ka and pKa

For a weak acid HA:

$$\mathrm{HA}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{A}}^{-}{K}_a=\frac{[{\mathrm{H}}_3{\mathrm{O}}^{+}][{\mathrm{A}}^{-}]}{[\mathrm{HA}]}$$

$$\mathrm{p}{K}_a=-{\log }_{10}{K}_a$$

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

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.
• $\mathrm{pH}=-\log [{\mathrm{H}}_3{\mathrm{O}}^{+}]$, $\mathrm{pH}+\mathrm{pOH}=14$, $K_w=1.0\times 10^{-14}$ 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