Acid strength, pH and the ionic product of water

What you’ll learn

• Identify conjugate acid–base pairs.
• Calculate pH for strong acids, strong bases and weak acids.
• State the assumptions in weak-acid calculations.

Understand the idea

A Brønsted–Lowry acid donates a proton and a base accepts one. A conjugate pair differs by exactly one proton. Strength describes the extent of ionisation; concentration describes amount per volume. A weak acid may be concentrated, and a strong acid may be dilute. Salt solutions can also be acidic or alkaline because their ions react with water.

For a strong monoprotic acid away from extreme dilution, hydrogen-ion concentration is approximately its analytical concentration. For a weak monoprotic acid, assume water contributes negligible hydrogen ions and the acid dissociates only slightly, giving $[\mathrm{H^+}]\approx\sqrt{K_ac}$. Use $K_w$ to calculate pH from a strong base. Neutrality means equal hydrogen- and hydroxide-ion concentrations; neutral pH is $7$ only when $\mathrm{p}K_w=14$, approximately at $25^\circ\mathrm C$.

Key formulae

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

$$K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]},\qquad\mathrm{p}K_a=-\log_{10}K_a$$

$$[\mathrm{H^+}]\approx\sqrt{K_ac}\quad\text{(weak monoprotic acid)}$$

$$K_w=[\mathrm{H^+}][\mathrm{OH^-}],\qquad\mathrm{pH}+\mathrm{pOH}=\mathrm{p}K_w$$

Worked example

Estimate the pH of a $0.10\,\mathrm{mol\,dm^{-3}}$ monoprotic acid with $K_a=1.0\times10^{-5}\,\mathrm{mol\,dm^{-3}}$.

1. Assume small dissociation and negligible ionisation of water.
2. Then $[\mathrm{H^+}]\approx\sqrt{10^{-5}\times0.10}=10^{-3}\,\mathrm{mol\,dm^{-3}}$.
3. The pH is $3.00$; dissociation is about $1\%$, supporting the approximation.

Answer: $\mathrm{pH}\approx3.00$.