Theory: Ionic Equilibrium
1. Henderson-Hasselbalch Equation
pH = pKa + log([A⁻]/[HA]). This is the fundamental buffer equation. At [A⁻] = [HA]: log(1) = 0 → pH = pKa. This is the half-equivalence point. The buffer has maximum capacity (resists pH change best) when pH = pKa. Buffer works best within pH = pKa ± 1.
2. Ka × Kb = Kw Relationship
For a conjugate acid-base pair: Ka(acid) × Kb(conjugate base) = Kw = 10⁻¹⁴ at 25°C. Given Kb(X⁻) = 10⁻¹⁰ → Ka(HX) = 10⁻¹⁴/10⁻¹⁰ = 10⁻⁴. This relationship is crucial for calculations involving weak acids and their conjugate bases.
3. Buffer Solutions and Buffer Capacity
A buffer resists pH change. Composed of weak acid + conjugate base (or weak base + conjugate acid). Buffer capacity: maximum when [acid] = [base] = pKa. Buffer range: pH = pKa ± 1. Blood pH 7·35–7·45 maintained by H₂CO₃/HCO₃⁻ (pKa=6·1) + other buffers. Adding strong acid to buffer: acid neutralised by A⁻ → little pH change. Adding strong base: neutralised by HA → little pH change.
4. Calculating pH of Buffer
pH = pKa + log([salt]/[acid]). Equivalently: pOH = pKb + log([salt]/[base]) for basic buffers. Note: if asked for pOH, use pOH = pKb + log([X⁻]/[X, where X is base]). In this problem, X⁻ IS the conjugate base; HX is the weak acid. Kb is for X⁻ (base). Ka is for HX (acid). Calculate Ka first, then use H-H equation.
5. Common Mistake: Confusing Ka and Kb
Kb is given for X⁻ (the base). The conjugate acid is HX. Ka(HX) = Kw/Kb(X⁻). THEN use Henderson-Hasselbalch with Ka(HX). Common error: using pKb directly in H-H equation → gives pOH not pH. Always identify: is the given K for the acid or the base? Then find the appropriate Ka for H-H.
6. Degree of Hydrolysis
Salts of weak acid + strong base: X⁻ + H₂O ⇌ HX + OH⁻, hydrolysis constant Kh = Kw/Ka = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰ = Kb(X⁻). Degree of hydrolysis h = √(Kh/C). pH of salt solution: pH = 7 + ½(pKa + log C). This connects buffer chemistry to salt hydrolysis.
7. Solubility and pH
pH affects solubility of sparingly soluble salts. For metal hydroxides: M(OH)n more soluble in acid (H⁺ consumes OH⁻ → shifts equilibrium right). For salts of weak acids (like CaCO₃): more soluble in acid (H⁺ + CO₃²⁻ → HCO₃⁻ → H₂CO₃ → CO₂ + H₂O → equilibrium shifts right). This is why caves (CaCO₃) form from acidic rain.
8. Polyprotic Acids
Have multiple Ka values. H₂SO₄: Ka₁ = very large (strong), Ka₂ = 0·012. H₂CO₃: Ka₁ = 4·3×10⁻⁷, Ka₂ = 4·8×10⁻¹¹. H₃PO₄: Ka₁ = 7·5×10⁻³, Ka₂ = 6·2×10⁻⁸, Ka₃ = 2·14×10⁻¹³. For buffering: each Ka gives one buffer range. Phosphate buffer (H₂PO₄⁻/HPO₄²⁻) at pH 7·2 is important in biology.
Frequently Asked Questions
1. Why is pH=pKa when [X⁻]=[HX]? ⌄
Henderson-Hasselbalch: pH = pKa + log([X⁻]/[HX]). When [X⁻]=[HX]: [X⁻]/[HX] = 1 → log(1) = 0 → pH = pKa + 0 = pKa. This is the half-equivalence point in a titration (halfway to equivalence, half the weak acid converted to conjugate base). The buffer has MAXIMUM buffering capacity here.
2. Why is pH=4 (acidic) if X⁻ is a base with Kb=10⁻¹⁰? ⌄
Ka(HX) = 10⁻⁴ → pKa = 4. HX is a moderately weak acid (pKa=4 < 7 → acidic). The buffer consisting of HX/X⁻ at equal concentrations has pH = pKa = 4. Even though X⁻ is a base, the equal mixture buffers at pH 4 (acidic side). For pH > 7 buffer, we'd need a weak acid with pKa > 7.
3. What if the buffer had [X⁻]=10[HX]? ⌄
pH = pKa + log(10/1) = 4 + 1 = 5. More base → higher pH. What if [HX]=10[X⁻]? pH = pKa + log(1/10) = 4 + (−1) = 3. More acid → lower pH. Henderson-Hasselbalch allows quick calculation of any buffer composition.
4. How does buffer capacity change with pH? ⌄
Maximum buffer capacity at pH = pKa (when [A⁻]=[HA]). At pH = pKa + 1: [A⁻]/[HA] = 10 (90% base, 10% acid) → less capacity to resist base addition. At pH = pKa − 1: [A⁻]/[HA] = 0·1 → less capacity to resist acid addition. Buffer works in range pKa ± 1. Outside this range, buffer capacity drops dramatically.
5. What is the significance of pKa = 4 in chemistry? ⌄
Organic acids with pKa ≈ 4: acetic acid (4·75), formic acid (3·74), benzoic acid (4·20), aspirin (3·5). Buffer range 3–5. Important in: food preservation (acetic acid in vinegar), pharmaceutical stability, protein chemistry (acidic amino acids: aspartic pKa=3·86, glutamic pKa=4·32).
6. Differentiate between buffer pH and equivalence point pH ⌄
At equivalence point (complete neutralisation of HX with strong base): only X⁻ remains → pH = 7 + ½(pKa + log C) > 7 (basic). Buffer pH (pKa=4) is the midpoint pH during titration (before equivalence). Equivalence pH for HX (pKa=4): pH = 7 + ½(4 + log C) ≈ 9 (if C=0.1M). These are very different! Buffer at pH=4; equivalence at pH≈9.
7. What is the Van Slyke equation? ⌄
Buffer capacity β = 2·303 C × Ka[H⁺]/(Ka+[H⁺])². Maximum at [H⁺] = Ka (i.e., pH = pKa). C = total buffer concentration. Higher total concentration → higher buffer capacity. This shows that diluting a buffer reduces its capacity (even though pH stays constant per H-H equation). The Van Slyke equation quantifies how well a buffer resists pH change.
8. How to select a buffer for a given pH? ⌄
Choose a weak acid/base pair with pKa within ±1 of desired pH. Examples: pH=3·74 → formate buffer (pKa=3·74). pH=4·75 → acetate buffer (pKa=4·75). pH=7·21 → phosphate buffer (pKa=7·21). pH=9·26 → ammonia buffer (pKa=9·26). pH=10·33 → carbonate buffer (pKa=10·33). In biology: intracellular phosphate buffer (pH≈7·2), bicarbonate buffer in blood (pH=7·4).