Acids, Bases, and pH

The pH Scale

Acidity is quantified by hydrogen-ion concentration:

$\text{pH} = -\log[H^+] \qquad \text{pOH} = -\log[OH^-]$

At 25 °C, water self-ionizes so that [H⁺][OH⁻] = Kw = 1.0 × 10⁻¹⁴, giving pH + pOH = 14. Because the scale is logarithmic, each pH unit is a 10× change in [H⁺]: a pH-2 solution is 1,000× more acidic than pH 5.

Acid–Base Definitions

Three definitions appear on the exam. An Arrhenius acid produces H⁺ in water and an Arrhenius base produces OH⁻. The broader Brønsted–Lowry definition calls an acid a proton (H⁺) donor and a base a proton acceptor, which introduces the idea of conjugate pairs: when HA donates a proton it becomes its conjugate base A⁻. The most general Lewis definition calls an acid an electron-pair acceptor and a base an electron-pair donor. For most FE problems the Brønsted–Lowry view is sufficient — track where the proton goes.

Strong Acids and Bases

Strong acids dissociate completely: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄. For a monoprotic strong acid, [H⁺] = the acid concentration.

• Worked example: 0.001 M HCl → [H⁺] = 10⁻³ → pH = −log(10⁻³) = 3.0.

Strong bases also fully dissociate: NaOH, KOH, Ca(OH)₂, Ba(OH)₂. For NaOH, [OH⁻] = concentration; for Ca(OH)₂ multiply by 2 (two OH⁻ per formula unit).

• Worked example: 0.001 M NaOH → [OH⁻] = 10⁻³ → pOH = 3.0 → pH = 14 − 3 = 11.0.

Trap: A 0.001 M Ca(OH)₂ solution has [OH⁻] = 0.002 M, so pOH = 2.70 and pH = 11.30 — don't forget the factor of 2.

Weak Acids and Bases

Weak acids/bases dissociate only partially. For HA ⇌ H⁺ + A⁻:

$K_a = \frac{[H^+][A^-]}{[HA]}$

Smaller Ka means weaker. For a conjugate acid–base pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴. To estimate [H⁺] of a weak acid of concentration C, use [H⁺] ≈ √(Ka·C) (valid when dissociation < ~5%).

• Worked example: 0.10 M acetic acid (Ka = 1.8 × 10⁻⁵): [H⁺] ≈ √(1.8×10⁻⁵ × 0.10) = √(1.8×10⁻⁶) = 1.34×10⁻³ → pH = −log(1.34×10⁻³) ≈ 2.87.

Buffer Solutions

A buffer resists pH change on adding small amounts of acid/base. It contains a weak acid and its conjugate base (e.g., CH₃COOH / CH₃COO⁻) or a weak base and its conjugate acid (NH₃ / NH₄⁺). The pH is given by the Henderson–Hasselbalch equation:

$\text{pH} = \text{p}K_a + \log\frac{[A^-]}{[HA]}$

• Worked example: 0.10 M acetic acid (Ka = 1.8 × 10⁻⁵, pKa = 4.74) + 0.15 M sodium acetate: pH = 4.74 + log(0.15/0.10) = 4.74 + 0.18 = 4.92.

Buffer capacity peaks when [A⁻] = [HA], at which point the log term is zero and pH = pKa. Choose a buffer whose pKa is within ±1 of the target pH. Biological and process buffers (phosphate, bicarbonate, acetate) all exploit this principle, and adding a strong acid simply converts some A⁻ back to HA without a large pH swing until the buffer is exhausted.

Weak Base Example

Weak bases work symmetrically. For ammonia, NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ with Kb = 1.8 × 10⁻⁵. For a 0.10 M NH₃ solution, [OH⁻] ≈ √(Kb·C) = √(1.8×10⁻⁵ × 0.10) = 1.34×10⁻³ M, so pOH = 2.87 and pH = 14 − 2.87 = 11.13. Note the symmetry with the acetic-acid acid example: identical Ka and Kb magnitudes give pH values that are mirror images about 7. The relation Ka × Kb = Kw lets you convert a weak acid's Ka into its conjugate base's Kb (and vice versa) — for the acetate/acetic-acid pair, Kb(acetate) = 10⁻¹⁴/(1.8×10⁻⁵) = 5.6 × 10⁻¹⁰.

Neutralization and Titration

Acid + base → salt + water. At the equivalence point, moles H⁺ = moles OH⁻:

$M_a V_a = M_b V_b$

• Worked example: mL of 0.50 M NaOH to neutralize 25 mL of 0.30 M HCl: V_b = (0.30 × 25)/0.50 = 15 mL.

For polyprotic acids (H₂SO₄, H₃PO₄), multiply the acid concentration by the number of acidic protons when finding equivalence.

Worked Titration Walkthrough

Titrate 50.0 mL of 0.100 M acetic acid (Ka = 1.8 × 10⁻⁵, pKa = 4.74) with 0.100 M NaOH. Initial moles of acid = 0.0500 L × 0.100 M = 5.00 × 10⁻³ mol. By M_aV_a = M_bV_b, the equivalence point needs V_b = (0.100 × 50.0)/0.100 = 50.0 mL of NaOH.

• Half-equivalence (25.0 mL added): half the acid is converted to acetate, so [A⁻] = [HA] and pH = pKa = 4.74. This is the standard way to read a weak acid's pKa straight off a titration curve.
• Equivalence point (50.0 mL added): all acid is now acetate. The solution is not neutral — acetate (the conjugate base of a weak acid) hydrolyzes water, giving a pH above 7 (typically ≈ 8.7 here once dilution to 100 mL total is accounted for).

The table below summarizes equivalence-point pH by titration class:

A suitable indicator changes color near the equivalence pH: phenolphthalein (≈ 8–10) suits weak-acid/strong-base titrations, while methyl orange (≈ 3–4.5) suits strong-acid/weak-base titrations. Choosing the indicator to match the equivalence pH minimizes titration error.