IB Chemistry R3.1 R3.1.3
R3.1.3

pH Calculations & Titrations

Core pH Equations

\( \text{pH} = -\log_{10}[\text{H}^+] \qquad [\text{H}^+] = 10^{-\text{pH}} \)

\( \text{pOH} = -\log_{10}[\text{OH}^-] \qquad \text{pH} + \text{pOH} = 14 \text{ (at 25°C)} \)

\( K_w = [\text{H}^+][\text{OH}^-] = 1.00 \times 10^{-14} \text{ at 25°C} \)

Worked Example

Find the pH of 0.020 mol dm⁻³ NaOH

NaOH is a strong base: [OH⁻] = 0.020 mol dm⁻³

pOH = −log(0.020) = 1.70

pH = 14 − 1.70 = 12.30

Titration Curves

Strong Acid + Strong Base Titration

Volume of base added / cm³ pH 0 7 14 Equivalence point (pH = 7 for SA/SB) Buffer region (HL: half-equivalence pH = pKa)

Buffer Solutions (HL)

A buffer resists changes in pH when small amounts of acid or base are added. Made from a weak acid + its conjugate base (or vice versa).

At the half-equivalence point: \( \text{pH} = pK_a \)

Think About It

Why is the equivalence point pH = 7 for strong acid/strong base but NOT 7 for weak acid/strong base?

At the equivalence point of a weak acid/strong base titration, the solution contains the conjugate base (e.g. CH₃COO⁻), which hydrolyses in water to produce OH⁻. This makes the solution basic (pH > 7).

← R3.1.2 Strong & WeakFlashcards →