Acids and bases

pH measures the hydrogen-ion concentration:

$\text{pH} = -\lg[\text{H}^+] \qquad [\text{H}^+] = 10^{-\text{pH}}$

Water self-ionises slightly, giving the ionic product of water:

$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\,\text{mol}^2\,\text{dm}^{-6}\;(298\,\text{K})$

Strong acids are fully dissociated, so [H⁺] equals the acid concentration. For 0.10 mol dm⁻³ HCl, pH = −lg(0.10) = 1.00.

Strong bases are fully dissociated to OH⁻; find [H⁺] via Kw. For 0.10 mol dm⁻³ NaOH: $[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.10} = 1.0 \times 10^{-13} \Rightarrow \text{pH} = 13.0$

A weak acid only partially dissociates, an equilibrium:

$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \qquad K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$

A larger Ka (or smaller pKa, where $\text{p}K_a = -\lg K_a$) means a stronger weak acid. Because dissociation is small, we approximate $[\text{H}^+] = [\text{A}^-]$ and $[\text{HA}] \approx$ the initial concentration, giving:

$[\text{H}^+] = \sqrt{K_a \times c}$

Worked. 0.10 mol dm⁻³ ethanoic acid, Ka = 1.8 × 10⁻⁵: $[\text{H}^+] = \sqrt{(1.8 \times 10^{-5})(0.10)} = 1.34 \times 10^{-3} \Rightarrow \text{pH} = 2.87$

Note this weak acid (pH 2.87) is far less acidic than 0.10 mol dm⁻³ HCl (pH 1.00) at the same concentration — strength (dissociation) is not the same as concentration.

A buffer resists changes in pH when small amounts of acid or alkali are added. An acidic buffer is a weak acid plus its conjugate base (e.g. ethanoic acid + sodium ethanoate). Rearranging Ka:

$[\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} \qquad \text{(or } \text{pH} = \text{p}K_a + \lg\frac{[\text{A}^-]}{[\text{HA}]}\text{)}$

It is the ratio of acid to conjugate base — not the absolute amounts — that sets the pH.

How it works — the buffer holds a large reservoir of both species: $\text{Added H}^+\!: \text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$ $\text{Added OH}^-\!: \text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$ so [H⁺] (and pH) barely changes.

Worked. 0.20 mol dm⁻³ acid + 0.10 mol dm⁻³ ethanoate, Ka = 1.8 × 10⁻⁵: $[\text{H}^+] = (1.8 \times 10^{-5})\frac{0.20}{0.10} = 3.6 \times 10^{-5} \Rightarrow \text{pH} = 4.44$

Indicators. Choose an indicator whose colour-change range lies within the steep (vertical) part of the titration curve — e.g. phenolphthalein for a weak acid–strong base titration (alkaline equivalence point), methyl orange for strong acid–weak base.