Why Boron Trifluoride Is an Acid Even Though It Has No Hydrogen

Lemon juice tastes sour, soap feels slippery and baking soda fizzes with vinegar. Chemists explain these behaviours with three definitions of acids and bases, each broader than the last, and measure acidity on the pH scale.

This lesson covers the Arrhenius, Bronsted-Lowry and Lewis concepts, pH calculations for strong and weak electrolytes, and Ostwald's dilution law.

Arrhenius acids release hydrogen ions and bases release hydroxide ions in water, Bronsted-Lowry acids donate protons and bases accept them, and Lewis acids accept an electron pair while Lewis bases donate one.

Arrhenius concept:

- Acid — gives ${\mathrm{H}}^{+}$ ions, present as ${\mathrm{H}}_3{\mathrm{O}}^{+}$, in water; hydrochloric acid
- Base — gives $\mathrm{O}{\mathrm{H}}^{-}$ ions in water; sodium hydroxide
- Limitation — applies only to water and cannot explain why ammonia is a base

Bronsted-Lowry concept:

- Acid — proton donor; base — proton acceptor
- Every acid-base reaction forms a conjugate acid and a conjugate base, which differ by one proton

Worked example. In $\mathrm{N}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{N}{\mathrm{H}}_4^{+}+\mathrm{O}{\mathrm{H}}^{-}$, water donates a proton, so it is the acid and $\mathrm{O}{\mathrm{H}}^{-}$ is its conjugate base; ammonia accepts the proton, so it is the base and $\mathrm{N}{\mathrm{H}}_4^{+}$ is its conjugate acid.

Amphoteric behaviour. Water acts as a base towards hydrochloric acid and as an acid towards ammonia.

Lewis concept:

- Lewis acid — electron-pair acceptor; $\mathrm{B}{\mathrm{F}}_3$, $\mathrm{AlC}{\mathrm{l}}_3$, ${\mathrm{H}}^{+}$ and $\mathrm{A}{\mathrm{g}}^{+}$
- Lewis base — electron-pair donor; $\mathrm{N}{\mathrm{H}}_3$, ${\mathrm{H}}_2\mathrm{O}$ and $\mathrm{O}{\mathrm{H}}^{-}$
- It covers reactions with no protons at all, such as boron trifluoride accepting the lone pair of ammonia

An everyday example. Antacid tablets taken for acidity after a spicy meal contain bases that neutralise excess stomach acid.

The substance. A strong acid has a weak conjugate base — hydrochloric acid gives up its proton so readily that the chloride ion barely takes it back.

**pH is the negative logarithm of hydrogen ion concentration, $\mathrm{pH}=-\log [{\mathrm{H}}^{+}]$, with $\mathrm{pH}+\mathrm{pOH}=14$ at 298 K; strong electrolytes ionise completely, while weak electrolytes ionise only partly, to an extent measured by the degree of ionisation $\alpha$.

Key relations at 298 K:**

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

Strong electrolytes ionise completely.

Worked example 1. In 0.010 M hydrochloric acid, $[{\mathrm{H}}^{+}]=1.0\times 10^{-2}$ M, so pH $=2$.

Worked example 2. In 0.0010 M sodium hydroxide, $[\mathrm{O}{\mathrm{H}}^{-}]=1.0\times 10^{-3}$ M, so pOH $=3$ and pH $=14-3=11$.

Weak electrolytes ionise partly. For a weak acid of concentration C:

$$K_a=\frac{C{\alpha }^2}{1-\alpha }\approx C{\alpha }^2[{\mathrm{H}}^{+}]=C\alpha =\sqrt{K_{a}C}$$

Worked example 3. For 0.10 M acetic acid, $K_a=1.8\times 10^{-5}$:

$$\alpha =\sqrt{\frac{1.8\times 10^{-5}}{0.10}}=0.0134[{\mathrm{H}}^{+}]=0.10\times 0.0134=1.34\times 10^{-3}\text{ M}$$

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

So only about 1.34 per cent of the acetic acid molecules are ionised.

Worked example 4 — a weak base. For 0.10 M ammonia with $K_b=1.8\times 10^{-5}$, $[\mathrm{O}{\mathrm{H}}^{-}]=1.34\times 10^{-3}$ M, so pOH $=2.87$ and pH $=11.13$.

An everyday example. The vinegar used in pickles is far less acidic than hydrochloric acid of the same concentration, because acetic acid ionises so little.

The substance. A very dilute acid can never have a pH above 7 — for $1.0\times 10^{-8}$ M hydrochloric acid, the hydrogen ions from water itself must be added, giving a pH of about 6.98, not 8.