QCE Chemistry - Unit 3 - Chemical equilibrium systems

pH | QCE Chemistry

Learn how pH links hydrogen ion concentration, acidity and logarithmic scale reasoning in QCE Chemistry.

Updated 2026-05-17 - 3 min read

pH is a logarithmic measure of acidity. More precisely, it is based on the concentration of hydrogen ions, usually represented in school chemistry as $[\mathrm{H^+}]$ or more accurately as hydronium ions, $[\mathrm{H_3O^+}]$.

Core formulas

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

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

$ \mathrm{pOH} = -\log_{10}[\mathrm{OH^-}] $

$ [\mathrm{OH^-}] = 10^{-\mathrm{pOH}} $

At $25^\circ\mathrm{C}$:

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

$ \mathrm{pH} + \mathrm{pOH} = 14 $

Acidic, neutral and basic

At $25^\circ\mathrm{C}$:

• acidic: $\mathrm{pH} < 7$, so $[\mathrm{H^+}] > [\mathrm{OH^-}]$
• neutral: $\mathrm{pH} = 7$, so $[\mathrm{H^+}] = [\mathrm{OH^-}]$
• basic: $\mathrm{pH} > 7$, so $[\mathrm{OH^-}] > [\mathrm{H^+}]$

Neutral means equal hydrogen ion and hydroxide ion concentrations. At $25^\circ\mathrm{C}$ that corresponds to pH 7, but the neutral pH can shift slightly at other temperatures because $K_w$ changes.

Self-ionisation of water

Water can react with itself:

$ 2\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{OH^-(aq)} $

This equilibrium exists even in pure water, but only to a tiny extent. That is why pure water has low conductivity compared with ionic solutions.

Strong acids and strong bases

For strong monoprotic acids, the hydrogen ion concentration is usually the same as the acid concentration because the acid is assumed to fully ionise. For example, $0.0100\ \mathrm{mol\ L^{-1}}$ hydrochloric acid gives:

$ [\mathrm{H^+}] = 0.0100\ \mathrm{mol\ L^{-1}} $

For strong bases, start with $[\mathrm{OH^-}]$, calculate pOH, then convert to pH:

$ \mathrm{pOH} = -\log_{10}[\mathrm{OH^-}] $

$ \mathrm{pH} = 14 - \mathrm{pOH} $

This is where students often rush. If the question gives a base, do not put $[\mathrm{OH^-}]$ directly into the pH formula. Use pOH first unless you have already converted to $[\mathrm{H^+}]$ using $K_w$.

Dilution and logarithms

Diluting an acid lowers $[\mathrm{H^+}]$, so pH increases. Diluting a base lowers $[\mathrm{OH^-}]$, so pOH increases and pH decreases. A tenfold dilution changes pH by 1 unit only for simple strong acid/base cases where the concentration changes cleanly by a factor of 10.

Worked examples

Comparing pH values

pH 2 has 100 times the hydrogen ion concentration of pH 4 because there are two pH units between them:

$ 10 \times 10 = 100 $

This kind of comparison is common in short-response questions.

Exam traps

Other traps:

• dropping the negative sign in $-\log_{10}[\mathrm{H^+}]$
• using mL instead of concentration in the pH formula
• rounding pH too early during multi-step calculations
• assuming neutral is always pH 7 without checking temperature context

Quick check

Sources

• QCAA Chemistry subject page
• QCAA Chemistry 2025 syllabus
• OpenStax Chemistry 2e: pH and pOH
• OpenStax Chemistry 2e: Relative Strengths of Acids and Bases