Appearance
10.5 — Acids, Bases and Electrochemistry
Your blood is held at pH 7.40, and if it moves outside 7.0 to 7.7 you die. A change of 0.4 in pH is fatal, and pH is logarithmic, so that corresponds to a factor of 2.5 in hydrogen ion concentration.
Meanwhile you eat acidic food, produce carbon dioxide continuously, and generate lactic acid whenever you exercise. Something is holding the line to two decimal places against constant assault, and this chapter explains what.
It also covers the other half of the same story: reactions that move electrons rather than protons, which is where batteries, corrosion and metabolism all live.
Three definitions of an acid
Arrhenius, 1884. An acid releases H⁺ in water; a base releases OH⁻. Simple and too narrow — it cannot explain why ammonia is basic, since NH₃ contains no OH.
Brønsted–Lowry, 1923. An acid is a proton donor; a base is a proton acceptor.
\text{HCl}+\text{H}_2\text{O} \to \text{H}_3\text{O}^++\text{Cl}^-
This covers ammonia:
\text{NH}_3+\text{H}_2\text{O} \rightleftharpoons \text{NH}_4^++\text{OH}^-
Ammonia accepts a proton, so it is a base. And it works without water at all, which the Arrhenius definition cannot.
Conjugate pairs. Every acid has a conjugate base, differing by one proton. A strong acid has a weak conjugate base — HCl gives Cl⁻, which has essentially no tendency to take a proton back.
Lewis, 1923. An acid is an electron pair acceptor; a base is a donor.
This is the most general. BF₃ has an incomplete octet (Chapter 10.1) and accepts a lone pair, so it is a Lewis acid with no protons involved at all. Every metal ion in solution is a Lewis acid, accepting lone pairs from water molecules, which is why metal ions are hydrated.
Use whichever fits. Brønsted–Lowry for aqueous chemistry, Lewis for coordination and organic mechanisms.
Water's autoionisation
Water reacts with itself:
2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^++\text{OH}^-
\boxed{K_w = [\text{H}^+][\text{OH}^-] = 1.0\times10^{-14}\ \text{at }25\ ^\circ\text{C}}
In pure water the two are equal:
[\text{H}^+] = \sqrt{10^{-14}} = 1.0\times10^{-7}\ \text{M}
How much water is ionised? Water is 55.5 M, so:
\frac{10^{-7}}{55.5} = 1.8\times10^{-9}
About two molecules in a billion. And yet that tiny fraction controls the acidity of everything.
K_w depends on temperature, because autoionisation is endothermic. At 100 °C, K_w = 5.1\times10^{-13}, so neutral pH is 6.14. Neutral does not mean pH 7; it means [\text{H}^+] = [\text{OH}^-], and the number depends on temperature.
pH
\boxed{\text{pH} = -\log_{10}[\text{H}^+], \qquad \text{pOH} = -\log_{10}[\text{OH}^-]}
And taking logs of K_w:
\text{pH}+\text{pOH} = 14
The scale is logarithmic, which is constantly underappreciated:
| pH | [\text{H}^+] | Example |
|---|---|---|
| 0 | 1 M | Battery acid |
| 1 | 10^{-1} | Stomach acid |
| 2.5 | 3\times10^{-3} | Lemon juice |
| 4.5 | 3\times10^{-5} | Acid rain |
| 5.6 | 2.5\times10^{-6} | Normal rain |
| 6.5 | 3\times10^{-7} | Milk |
| 7.4 | 4\times10^{-8} | Blood |
| 8.1 | 8\times10^{-9} | Seawater |
| 11 | 10^{-11} | Ammonia solution |
| 14 | 10^{-14} | Drain cleaner |
Stomach acid at pH 1 is a million times more acidic than blood at pH 7.4.
And normal rain is pH 5.6, not 7 — dissolved atmospheric CO₂ forms carbonic acid. Acid rain means below about 5.0, from sulphur and nitrogen oxides.
Strong acids
Fully dissociated, so the calculation is trivial.
Worked example: 0.010 M HCl.
[\text{H}^+] = 0.010\ \text{M} \quad\Longrightarrow\quad \text{pH} = -\log(0.010) = 2.00
A trap worth knowing. For 10^{-8} M HCl, naively pH = 8 — which would make an acid basic. The error is ignoring water's own 10^{-7} M, which now dominates. The correct treatment gives pH 6.98.
Weak acids
Partially dissociated, and the extent is given by K_a.
\text{HA} \rightleftharpoons \text{H}^++\text{A}^-, \qquad K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
\text{p}K_a = -\log K_a
| Acid | K_a | p$K_a$ |
|---|---|---|
| Hydrofluoric | 6.8\times10^{-4} | 3.17 |
| Formic | 1.8\times10^{-4} | 3.75 |
| Acetic | 1.8\times10^{-5} | 4.76 |
| Carbonic (1st) | 4.5\times10^{-7} | 6.35 |
| Dihydrogen phosphate | 6.2\times10^{-8} | 7.21 |
| Ammonium | 5.6\times10^{-10} | 9.25 |
Worked example: 0.100 M acetic acid.
Set up the table with x as the amount dissociated:
K_a = \frac{x^2}{0.100-x} = 1.8\times10^{-5}
Assume x \ll 0.100, which is valid when K_a is small:
x^2 = 1.8\times10^{-6} \quad\Longrightarrow\quad x = 1.34\times10^{-3}
Check the assumption: 1.34\times10^{-3}/0.100 = 1.3 %, comfortably under the usual 5 % threshold. ✔
\text{pH} = -\log(1.34\times10^{-3}) = 2.87
Compare with 0.100 M HCl at pH 1.00. Same concentration, and the weak acid is 74 times less acidic.
Buffers
A buffer resists pH change, and it is a mixture of a weak acid and its conjugate base.
The mechanism. Add acid and the conjugate base mops it up:
\text{A}^-+\text{H}^+ \to \text{HA}
Add base and the acid neutralises it:
\text{HA}+\text{OH}^- \to \text{A}^-+\text{H}_2\text{O}
Either way the added species is consumed and the pH barely moves.
The Henderson–Hasselbalch equation
Start from the K_a expression:
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
Rearrange:
[\text{H}^+] = K_a\frac{[\text{HA}]}{[\text{A}^-]}
Take -\log of both sides:
\boxed{\text{pH} = \text{p}K_a+\log\frac{[\text{A}^-]}{[\text{HA}]}}
Read it: the pH is set by the p$K_a$ and adjusted by the log of the ratio. When the two are equal, pH = p$K_a$ exactly.
And the buffer works best there, because the ratio changes most slowly when it is near 1. Useful buffering range is p$K_a \pm 1$.
Worked example: how well does a buffer actually work?
A buffer contains 0.100 mol acetic acid and 0.100 mol acetate in 1.00 L.
\text{pH} = 4.76+\log(1.00) = 4.76
Add 0.010 mol of HCl.
The acid converts acetate to acetic acid:
[\text{A}^-] = 0.090, \qquad [\text{HA}] = 0.110
\text{pH} = 4.76+\log\frac{0.090}{0.110} = 4.76+\log(0.818) = 4.76-0.087 = 4.67
pH changed by 0.09.
Now add the same acid to pure water at pH 7:
[\text{H}^+] = 0.010 \quad\Longrightarrow\quad \text{pH} = 2.00
pH changed by 5.00.
The buffer reduced the change by a factor of 55. That is what buffering means, quantitatively.
Blood
The bicarbonate buffer is the main one:
\text{CO}_2+\text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^++\text{HCO}_3^-
With p$K_a = 6.10$ and blood pH 7.40:
7.40 = 6.10+\log\frac{[\text{HCO}_3^-]}{[\text{CO}_2]} \quad\Longrightarrow\quad \frac{[\text{HCO}_3^-]}{[\text{CO}_2]} = 10^{1.30} = 20
A 20:1 ratio, which is far from the ideal 1:1 and should make it a poor buffer.
It is an excellent buffer for a reason no test tube can match: it is open. The CO₂ concentration is controlled by breathing and the bicarbonate by the kidneys, so both components are actively regulated.
Breathe faster and you blow off CO₂, shifting the equilibrium left, raising pH — respiratory alkalosis, and it is why hyperventilating causes tingling and dizziness, and why breathing into a bag (which re-inhales your own CO₂) corrects it.
Hold your breath and CO₂ accumulates, lowering pH — respiratory acidosis.
The kidneys adjust bicarbonate over hours to days, providing slow correction.
Two control systems on one buffer, one fast and one slow. That is why 7.40 is held to two decimal places.
Titration
Add base to acid and track the pH.
The half-equivalence point is where half the acid has been neutralised, so [\text{HA}] = [\text{A}^-]:
\text{pH} = \text{p}K_a
This is the standard way to measure K_a experimentally — titrate, find the half-equivalence point, read the pH.
The equivalence point is where the moles of base equal the moles of acid. For a strong acid with a strong base it is pH 7. For a weak acid with a strong base it is above 7, because the conjugate base left in solution is itself basic.
Worked example: acetic acid with NaOH. At the equivalence point you have 0.050 M sodium acetate. The acetate hydrolyses:
K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{1.8\times10^{-5}} = 5.6\times10^{-10}
[\text{OH}^-] = \sqrt{K_b\times0.050} = \sqrt{2.8\times10^{-11}} = 5.3\times10^{-6}
\text{pOH} = 5.28 \quad\Longrightarrow\quad \text{pH} = 8.72
Which is why the indicator matters. Phenolphthalein changes colour between pH 8.2 and 10 and is right for this titration. Methyl orange changes at 3.1 to 4.4 and would give a badly wrong endpoint.
Redox
The other half of chemistry: reactions that move electrons.
\text{Oxidation} = \text{loss of electrons}, \qquad \text{Reduction} = \text{gain of electrons}
The mnemonic is OIL RIG — oxidation is loss, reduction is gain.
Oxidation number rules, applied in order:
- Element in its standard state: 0.
- Monatomic ion: its charge.
- Fluorine: always -1.
- Hydrogen: +1 (except -1 in metal hydrides).
- Oxygen: -2 (except -1 in peroxides, +2 in OF₂).
- Sum equals the overall charge.
Worked example: the oxidation state of manganese in MnO₄⁻.
\text{Mn}+4(-2) = -1 \quad\Longrightarrow\quad \text{Mn} = +7
Manganese in its highest possible oxidation state, which is why permanganate is such a strong oxidiser — it has a great deal of room to gain electrons.
Balancing by half-reactions
Worked example, in acid:
\text{MnO}_4^-+\text{Fe}^{2+} \to \text{Mn}^{2+}+\text{Fe}^{3+}
Step 1: split into half-reactions.
\text{MnO}_4^- \to \text{Mn}^{2+} \qquad \text{Fe}^{2+} \to \text{Fe}^{3+}
Step 2: balance O with water, H with H⁺.
\text{MnO}_4^- \to \text{Mn}^{2+}+4\text{H}_2\text{O}
\text{MnO}_4^-+8\text{H}^+ \to \text{Mn}^{2+}+4\text{H}_2\text{O}
Step 3: balance charge with electrons. Left is -1+8 = +7; right is +2. Add 5 electrons to the left:
\text{MnO}_4^-+8\text{H}^++5e^- \to \text{Mn}^{2+}+4\text{H}_2\text{O}
Step 4: the other half.
\text{Fe}^{2+} \to \text{Fe}^{3+}+e^-
Step 5: equalise electrons and add. Multiply the iron half by 5:
\boxed{\text{MnO}_4^-+8\text{H}^++5\text{Fe}^{2+} \to \text{Mn}^{2+}+4\text{H}_2\text{O}+5\text{Fe}^{3+}}
Check: charge is -1+8+10 = +17 on the left and +2+15 = +17 on the right. ✔
This titration is self-indicating — permanganate is deep purple and Mn²⁺ is nearly colourless, so the endpoint is the first permanent pink.
Electrochemical cells
Separate the two half-reactions and make the electrons travel through a wire.
The Daniell cell:
\text{Anode (oxidation)}: \quad \text{Zn} \to \text{Zn}^{2+}+2e^-
\text{Cathode (reduction)}: \quad \text{Cu}^{2+}+2e^- \to \text{Cu}
The salt bridge is essential. Without it, the zinc side accumulates positive charge and the copper side negative, the potential difference builds up in opposition, and current stops within microseconds. The bridge lets ions migrate to keep both sides neutral.
Mnemonic: an ox (anode = oxidation), red cat (reduction = cathode). And electrons flow from anode to cathode through the external circuit, always.
Standard electrode potentials
Measured against the standard hydrogen electrode, defined as exactly 0 V.
| Half-reaction | E^\circ (V) |
|---|---|
| F₂ + 2e⁻ → 2F⁻ | +2.87 |
| MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | +1.51 |
| O₂ + 4H⁺ + 4e⁻ → 2H₂O | +1.23 |
| Ag⁺ + e⁻ → Ag | +0.80 |
| Cu²⁺ + 2e⁻ → Cu | +0.34 |
| 2H⁺ + 2e⁻ → H₂ | 0.00 |
| Fe²⁺ + 2e⁻ → Fe | -0.44 |
| Zn²⁺ + 2e⁻ → Zn | -0.76 |
| Al³⁺ + 3e⁻ → Al | -1.66 |
| Li⁺ + e⁻ → Li | -3.04 |
More positive means a stronger tendency to be reduced. Fluorine is the strongest oxidiser known; lithium is the strongest reducer.
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}
For the Daniell cell:
E^\circ = 0.34-(-0.76) = 1.10\ \text{V}
And the connection to thermodynamics:
\boxed{\Delta G^\circ = -nFE^\circ}
where F = 96{,}485 C/mol is the Faraday constant — the charge on a mole of electrons.
\Delta G^\circ = -(2)(96485)(1.10) = -212{,}000\ \text{J/mol} = -212\ \text{kJ/mol}
Positive cell voltage means negative \Delta G means spontaneous. The two ways of describing favourability are the same statement.
The Nernst equation
Standard potentials assume 1 M concentrations. Real cells are not at standard conditions, and the correction is the Nernst equation.
Derivation. From Chapter 10.6's relation for non-standard free energy:
\Delta G = \Delta G^\circ + RT\ln Q
Substitute \Delta G = -nFE throughout:
-nFE = -nFE^\circ + RT\ln Q
Divide by -nF:
\boxed{E = E^\circ - \frac{RT}{nF}\ln Q}
At 25 °C, converting to base-10 logs:
\frac{RT}{F}\ln 10 = \frac{(8.314)(298.15)}{96485}\times2.303 = 0.0592
\boxed{E = E^\circ - \frac{0.0592}{n}\log Q}
Worked example: a concentration cell. Two copper electrodes in solutions of 1.0 M and 0.001 M Cu²⁺.
Both half-reactions are identical, so E^\circ = 0 — and yet a voltage appears:
E = 0 - \frac{0.0592}{2}\log\frac{0.001}{1.0} = -0.0296\times(-3) = +0.089\ \text{V}
89 millivolts from nothing but a concentration difference. The driving force is entropy: the system moves towards equal concentrations (Chapter 3.5).
This is exactly how a pH meter works. A glass electrode develops a potential across a thin glass membrane that depends on the hydrogen ion concentration difference:
E = \text{const} - 0.0592\,\text{pH}
59.2 mV per pH unit at 25 °C — a number every pH meter is calibrated against, and it comes straight from RT\ln 10/F.
And it is how nerve impulses work. A neuron maintains a potassium gradient across its membrane, roughly 140 mM inside and 5 mM outside:
E = -\frac{0.0592}{1}\log\frac{5}{140} = -0.0592\times(-1.447) = -0.086\ \text{V}
About -86 mV, which is close to the measured resting potential of -70 mV. Every thought you have runs on the Nernst equation.
Batteries
| Type | Chemistry | Voltage |
|---|---|---|
| Alkaline | Zn / MnO₂ | 1.5 V |
| Lead-acid | Pb / PbO₂ | 2.1 V per cell |
| NiMH | Metal hydride / NiOOH | 1.2 V |
| Lithium-ion | Li in graphite / LiCoO₂ | 3.7 V |
Lithium-ion's advantage is threefold. Lithium is the lightest metal (6.94 g/mol), it has the most negative standard potential (-3.04 V), and the intercalation chemistry is reversible for hundreds of cycles.
It does not contain lithium metal. Lithium ions shuttle between graphite and metal oxide layers, sliding in and out of the lattice without destroying it. That reversibility is the whole invention, and Goodenough, Whittingham and Yoshino shared the 2019 Nobel Prize for it. Goodenough was 97, the oldest Nobel laureate ever.
Lead-acid's 2.1 V is, as Chapter 9.3 noted, about 1.7 V of relativistic effect on lead's 6s orbital.
Corrosion
\text{Fe} \to \text{Fe}^{2+}+2e^-, \qquad E^\circ = +0.44\ \text{V}
\text{O}_2+2\text{H}_2\text{O}+4e^- \to 4\text{OH}^-, \qquad E^\circ = +0.40\ \text{V}
E^\circ_{\text{cell}} = 0.84\ \text{V}
Strongly spontaneous. Iron in wet air is a short-circuited battery, and Chapter 9.6 explained why the product flakes off instead of protecting.
Three defences, all electrochemical:
Barrier — paint, plating, oil. Keeps out water and oxygen.
Sacrificial anode — bolt on a more easily oxidised metal. Zinc at -0.76 V oxidises preferentially and protects the iron. This is galvanising, and it is why ships and pipelines carry blocks of zinc or magnesium that are replaced periodically.
Impressed current — apply an external voltage to hold the structure negative. Used on large pipelines and bridges.
And the danger of mixing metals. Bolt copper (+0.34 V) to steel (-0.44 V) in a wet environment and you have built a cell with 0.78 V driving the steel's destruction. This is galvanic corrosion, and it is why marine fittings specify compatible metals and insulating washers.
Where this shows up in your life
Your blood pH, held to two decimals by a buffer with two independent control loops.
Every battery you own.
Antacids neutralise stomach acid; proton pump inhibitors block the pump that makes it.
Baking. Baking soda plus an acid releases CO₂; baking powder contains both.
Bleach is oxidation, breaking the conjugated double bonds that make coloured molecules coloured.
Photography, electroplating, aluminium smelting and chlorine production are all industrial electrochemistry.
Fuel cells run the hydrogen–oxygen reaction directly, at E^\circ = 1.23 V, without a flame.
And your own metabolism is a controlled redox cascade: glucose is oxidised, oxygen is reduced, and the electrons pass down a chain of proteins that pumps protons across a membrane to make ATP. Respiration is a battery, and Chapter 10.8 comes back to it.
What the next chapter fixes
Two questions have been answered separately and never connected. Why does a reaction go one way rather than the other, and what decides how fast? Chapter 10.6 gives both: it derives Gibbs free energy and shows why \Delta G < 0 is the criterion for spontaneity, explains how a reaction can be favourable and infinitely slow, derives the Arrhenius equation from the Maxwell–Boltzmann distribution of Chapter 3.2, and shows exactly what a catalyst does and does not do.