pH of Strong Acids & Bases SL
Strong acids and bases fully dissociate, so the initial concentration directly gives you [H⁺] or [OH⁻].
Essential pH and pOH Relationships
\[\text{pH} = -\log_{10}[\text{H}^+], \quad \text{pOH} = -\log_{10}[\text{OH}^-]\] \[[\text{H}^+] = 10^{-\text{pH}}, \quad [\text{OH}^-] = 10^{-\text{pOH}}\] \[\text{pH} + \text{pOH} = 14.00 \quad (\text{at } 298\text{ K})\]
pH of a Strong Acid
Problem: Calculate the pH of a \(0.10\text{ mol dm}^{-3}\text{ HCl}\) solution.
Solution: \(\text{HCl}\) is a strong monoprotic acid → \([\text{H}^+] = 0.10\text{ mol dm}^{-3}\).
\[\text{pH} = -\log_{10}(0.10) = 1.00\]pH of a Diprotic Strong Base
Problem: Calculate the pH of \(0.0044\text{ mol dm}^{-3}\text{ Ca(OH)}_2\) at \(298\text{ K}\).
Solution: \(\text{Ca(OH)}_2\) releases \(2\text{ OH}^-\) ions per formula unit:
\[[\text{OH}^-] = 2 \times 0.0044 = 0.0088\text{ mol dm}^{-3}\] \[\text{pOH} = -\log_{10}(0.0088) = 2.06\] \[\text{pH} = 14.00 - 2.06 = 11.94\]HLpH of Weak Acids (Using Ka)
Weak acids partially dissociate. Use an ICE table to find [H⁺].
Worked Example: pH of a Weak Acid
Problem: Calculate the pH of \(0.534\text{ mol dm}^{-3}\) formic acid (\(\text{HCOOH}\)), given \(K_{\text{a}} = 1.77 \times 10^{-4}\).
Solution: Using the small-\(x\) approximation (\([\text{H}^+] = \sqrt{K_{\text{a}} \times c}\)):
\[[\text{H}^+] = \sqrt{1.77 \times 10^{-4} \times 0.534} = \sqrt{9.45 \times 10^{-5}} = 9.72 \times 10^{-3}\text{ mol dm}^{-3}\] \[\text{pH} = -\log_{10}(9.72 \times 10^{-3}) = 2.01\]HLpH of Weak Bases (Using Kb)
Worked Example: pH of a Weak Base
Problem: Calculate the pH of \(0.0325\text{ mol dm}^{-3}\text{ NH}_3\), given \(K_{\text{b}} = 1.77 \times 10^{-5}\).
Solution: \([\text{OH}^-] = \sqrt{K_{\text{b}} \times c} = \sqrt{1.77 \times 10^{-5} \times 0.0325} = 7.58 \times 10^{-4}\text{ mol dm}^{-3}\)
\[\text{pOH} = -\log_{10}(7.58 \times 10^{-4}) = 3.12 \implies \text{pH} = 14.00 - 3.12 = 10.88\]Titration Curves
Strong Acid + Strong Base Titration
HLAll Four Titration Curve Types
| Combination | Initial pH | Equiv. PH | Key Features |
|---|---|---|---|
| SA + SB | ~1 | 7 | Steep symmetric jump; no buffer region |
| WA + SB | ~3-4 | > 7 | Buffer region before equiv.; half-equiv: pH = pKa |
| SA + WB | ~11 | < 7 | Buffer region; half-equiv: pOH = pKb |
| WA + WB | Moderate | Varies | Very shallow curve; no steep section; pH meter needed |
HLChoosing an Indicator
An indicator is a weak acid (HInd) whose conjugate base (Ind⁻) has a different colour. It changes colour when pH ≈ pKa of the indicator (±1).
Indicator Selection Rule
An indicator is suitable for a titration if its \(\text{p}K_{\text{In}} \pm 1\) range falls entirely within the vertical inflection jump of the pH curve.
- Strong Acid + Strong Base: Jump pH 3–11 → Phenolphthalein or Methyl orange.
- Strong Acid + Weak Base: Jump pH 3–7 → Methyl orange (\(\text{p}K_{\text{In}} \approx 3.7\)).
- Weak Acid + Strong Base: Jump pH 7–11 → Phenolphthalein (\(\text{p}K_{\text{In}} \approx 9.3\)).
- Weak Acid + Weak Base: No sharp vertical jump → No indicator suitable (use pH probe).
HLSalt Hydrolysis
When a salt dissolves, its ions may react with water (hydrolyse). The pH depends on the parent acid and base:
| Salt Type | Made From | Solution pH | Example |
|---|---|---|---|
| Neutral | SA + SB | = 7 | NaCl |
| Basic | WA + SB | > 7 | CH₃COONa |
| Acidic | SA + WB | < 7 | NH₄Cl |
HLBuffer Solutions
A buffer resists changes in pH when small amounts of acid or base are added. Made from comparable amounts of a conjugate pair.
Buffer Action and Henderson-Hasselbalch
Acidic Buffer: Weak acid + conjugate salt (e.g. \(\text{CH}_3\text{COOH} / \text{CH}_3\text{COONa}\)).
Basic Buffer: Weak base + conjugate salt (e.g. \(\text{NH}_3 / \text{NH}_4\text{Cl}\)).
Henderson-Hasselbalch Equation:
\[\text{pH} = \text{p}K_{\text{a}} + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]When \([\text{A}^-] = [\text{HA}]\), \(\text{pH} = \text{p}K_{\text{a}}\) (maximum buffer capacity).
Henderson-Hasselbalch Equation
pH = pKa + log([A⁻] / [HA])
At the half-equivalence point, [A⁻] = [HA], so log(1) = 0 → pH = pKa. This is how you experimentally determine Ka from a titration curve.
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