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Axioma

Chemistry · Advanced · 30 min · Chapter 26 of 28

Acids, bases, and electrochemistry

pH, weak-acid equilibria, the Henderson–Hasselbalch equation, and the voltage of an electrochemical cell.

Read first: Equilibrium, energy, and rates

The idea

Water autoionizes: H₂O ⇌ H⁺ + OH⁻, with K_w = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 25 °C. In pure water each ion is 1.0×10⁻⁷ M and the pH is 7. Acids raise [H⁺] and, because the product is fixed, lower [OH⁻]. A strong acid such as HCl is treated as fully dissociated, so 0.010 M HCl has pH 2.00. A weak acid is not. You solve the equilibrium Kₐ = [H⁺][A⁻]/[HA], often with the approximation that x is small compared with the initial concentration, and you check that approximation afterward.

A buffer is a weak acid and its conjugate base, both present in comparable amounts. Adding a little strong acid converts some base into the weak acid; adding a little strong base does the reverse. The pH barely moves. The Henderson–Hasselbalch equation reads the equilibrium in log form: pH = pKₐ + log([A⁻]/[HA]). When the two concentrations are equal, pH = pKₐ, which is the point where the buffer works best.

An electrochemical cell runs a redox reaction in two places joined by a wire and a salt bridge. The standard cell potential is E°_cathode − E°_anode, using reduction potentials for both and subtracting. A positive E° means a spontaneous reaction as written, under standard conditions. The free-energy connection is ΔG = −nFE, so a positive voltage is a negative free-energy change. The Nernst equation corrects the voltage for concentrations that are not standard. Electrolysis is the same apparatus run backwards by an external voltage.

Hold onto this

  1. pH is the negative log of the hydrogen-ion activity. In dilute water it is about −log[H⁺].
  2. A weak acid’s dissociation is an equilibrium with constant Kₐ. Strong acids are treated as fully dissociated.
  3. A buffer resists pH change. Its pH is set by the pKₐ and the ratio of conjugate base to acid.
  4. Cell voltage is a free-energy measurement: ΔG = −nFE.

Formulas

pH

At 25 °C, K_w = 1.0×10⁻¹⁴ and a neutral solution has pH 7.

[H⁺]
Hydrogen-ion concentration in mol/L

Acid dissociation

Equilibrium constant for HA ⇌ H⁺ + A⁻. Larger Kₐ means a stronger weak acid.

pKₐ
−log Kₐ

Henderson–Hasselbalch

pH of a buffer. The ratio is conjugate base over weak acid.

[A⁻]/[HA]
Base-to-acid ratio

Standard cell potential

Both potentials are reduction potentials. The anode is the electrode where oxidation occurs.

E°
Standard reduction potential

Free energy of a cell

Voltage and free energy describe the same reaction. F is the charge of one mole of electrons.

n
Moles of electrons transferred
F
Faraday constant, 96485 C/mol

Nernst equation

Cell voltage at nonstandard concentrations. At 25 °C the prefactor is about (0.059 V)/n when written with log base 10.

Q
Reaction quotient

Worked examples

1. A strong acid

What is the pH of 0.010 M HCl?

  1. HCl is strong, so [H⁺] ≈ 0.010 M.
  2. pH = −log(0.010) = 2.00.
  3. Check [OH⁻] = 10⁻¹⁴ / 0.010 = 10⁻¹², negligible as a correction to [H⁺].

pH 2.00.

2. A buffer at equal concentrations

Acetic acid has pKₐ = 4.76. What is the pH of a buffer with [acetate] = [acetic acid] = 0.10 M? What if you then double the acetate?

  1. When the ratio is 1, log 1 = 0, so pH = 4.76.
  2. Doubling acetate makes the ratio 2.
  3. log₁₀ 2 ≈ 0.30, so pH ≈ 5.06.

4.76, then about 5.06. Tenfold changes in the ratio move the pH by one unit.

Try these

A cell has E° = 0.46 V and transfers 2 electrons. Estimate ΔG°.

The pH is 3.50. What is [H⁺]?

Common mistakes

  • Henderson–Hasselbalch wants base over acid. Inverting the ratio changes the sign of the log and gives the wrong side of the pKₐ.
  • Strong acids do not have a meaningful Kₐ in the beginner treatment. Do not set up an ICE table for 0.01 M HCl.
  • The anode is where oxidation happens, which is not always the electrode drawn on the left. The sign of E° follows the chemistry, not the picture.