Imagine you want to know how a reaction's speed depends on the concentration of one particular reactant — iodine. The problem is, you can't just "watch" a concentration change directly. What you can do is repeatedly grab small samples of the reaction mixture at known times, instantly stop ("quench") each sample from reacting further, and then titrate it to work out exactly how much iodine was left at that moment. Plot those values against time, and the shape of the graph tells you the order of reaction.

The reaction studied here is the acid-catalysed iodination of propanone:

Both propanone and the H⁺ catalyst are used in large excess. This is the clever trick of the whole experiment: because there's so much of them compared to the iodine, their concentrations barely change over the course of the reaction, even though iodine's concentration is dropping steadily. That means any change in rate we observe can only be blamed on the changing iodine concentration — every other variable is effectively held constant. This is sometimes called the "isolation method."

1. Mix 25 cm³ of 1.0 mol dm⁻³ propanone with 25 cm³ of 1.0 mol dm⁻³ sulfuric acid in a beaker.
2. Add 50 cm³ of 0.02 mol dm⁻³ iodine solution — start the timer at this exact moment.
3. At regular intervals, pipette out a 10 cm³ portion of the mixture into a conical flask.
4. Quench the sample immediately by adding a spatula of sodium hydrogencarbonate. This neutralises the sulfuric acid catalyst, so the reaction in that particular sample stops dead — it's now a frozen snapshot of the reaction mixture at that exact time.
5. Titrate the quenched sample against 0.01 mol dm⁻³ sodium thiosulfate(VI) solution, using starch as the indicator, to find out how much iodine remains.
6. Record time and titre in a table, and repeat for several time points.

The titre volume of thiosulfate is directly proportional to the concentration of iodine remaining in the sample (via the reaction below), so a graph of titre vs time behaves exactly like a concentration-vs-time graph.

Plotting this data gives a straight line with a constant negative gradient — the titre falls at a steady rate the whole way through, never curving or levelling off.

A clock reaction is named for its party trick: nothing visible happens for a while, and then suddenly — at one precise instant — there's a sharp, dramatic colour change, like an alarm going off. This makes clock reactions brilliant for kinetics, because instead of continuously monitoring concentration, you just need a stopwatch and your eyes. You measure the time taken for the colour to appear, and that time is directly linked to the rate of reaction.

This experiment uses the reaction between hydrogen peroxide and iodide ions:

On its own, this reaction would just slowly turn the solution brown/black as iodine builds up in the presence of starch. To make it a "clock," we add a small, fixed amount of sodium thiosulfate, which reacts with iodine as fast as it's made:

Because there's only a small, known amount of thiosulfate, it gets completely used up at some point. Only after the thiosulfate runs out does iodine start to accumulate — and the moment it does, the starch indicator immediately flashes blue-black. That sudden colour change is the "clock" going off.

• All solutions except sulfuric acid (which is in large excess) are measured out in burettes for precision, and placed in a small beaker.
• The reaction is started by adding 1 cm³ of 0.25 mol dm⁻³ hydrogen peroxide and starting the timer at the same moment.
• The timer is stopped the instant the blue-black colour appears.
• Different runs use different volumes of KI (with water topping up the total volume), so [I⁻] varies systematically while everything else stays constant.

Here's the key conceptual leap in this practical: the time for the colour to appear (t) is inversely proportional to the rate of reaction. A fast reaction uses up the fixed thiosulfate quickly, so the colour appears sooner (small t); a slow reaction takes longer (large t). So instead of plotting t directly, we plot 1/t, which behaves just like "rate."

Plotting rate (1/t) against [KI] gives a straight line through the origin — rate is directly proportional to concentration. Doubling [KI] doubles the rate.

You already know that reactions go faster when you heat them up. But how much faster, and why, is governed precisely by the Arrhenius equation. This practical uses a clock reaction (similar in spirit to CP9b) run at several different temperatures to work out the rate constant k at each temperature, and then uses those k values to calculate the reaction's activation energy, Eₐ.

The reaction studied is between bromate(V) and bromide ions, catalysed by acid, which generates bromine:

Just like the thiosulfate "mops up" iodine in CP9b, here phenol reacts with the bromine as fast as it's produced, so no bromine actually accumulates while phenol is still present:

Once all the phenol has reacted, any further bromine produced is free to bleach the methyl red (or methyl orange) indicator, giving a sudden, clear endpoint.

1. Pipette 10.0 cm³ of phenol solution and 10.0 cm³ of the bromide/bromate solution into a boiling tube, and add methyl red indicator.
2. Pipette 5.0 cm³ of sulfuric acid into a separate boiling tube.
3. Stand both tubes in a water bath at a set temperature (e.g. 37 °C ± 1 °C) until their contents reach that temperature.
4. Mix the two tubes together rapidly (pour back and forth), and start the stop clock at the same instant.
5. Time how long it takes for the methyl red indicator to disappear (bleach).
6. Repeat the whole experiment at several different temperatures (ice can be used to reach lower temperatures than room temperature).

This is the trickiest maths in the whole chapter, so let's build it up piece by piece rather than just handing you the final graph.

Step 1 — Time to rate constant. Just like in CP9b, the time for the colour change is inversely related to rate, and in this simplified treatment, 1/t is taken as proportional to the rate constant k itself (since the concentrations of reactants are the same in each run, only temperature is varied).

Step 2 — The Arrhenius equation. This is the master formula linking k to temperature:

Step 3 — Turn it into a straight-line equation. Taking natural logs of both sides transforms this exponential relationship into something you can plot as a straight line — this is the trick that makes the whole method work:

Step 4 — Plot the graph. Plotting ln k against 1/T gives a straight line with a negative gradient (as temperature rises, 1/T falls, and ln k rises — consistent with reactions going faster when hotter).

Step 5 — Find the gradient. Using two well-spaced points on the line of best fit:

Step 6 — Calculate Eₐ. Since gradient = −Eₐ/R:

Step 7 — Calculate A (the pre-exponential factor), if asked. Pick any point that lies on your line of best fit (it doesn't need to be one of your original data points — read it straight off the graph). Substitute its ln k and 1/T values, plus the gradient you already found, into y = mx + c and solve for ln A, then take e^(ln A) to get A.

Ka measures how far a weak acid dissociates in water:

This experiment uses a beautifully simple trick to measure Ka without needing to know exact concentrations of everything. Imagine you take a flask of weak acid and titrate it with sodium hydroxide until it's exactly half neutralised — meaning half of the original acid molecules (HA) have been converted into their conjugate base (A⁻), and half remain as HA. At that specific point, [HA] and [A⁻] are equal, so they cancel out of the Ka expression entirely:

So all you need to do is measure the pH at the half-equivalence point, convert it to [H⁺], and that value is Ka. No further calculation needed!

1. Calibrate a pH probe/meter.
2. Pipette 25 cm³ of 0.1 mol dm⁻³ ethanoic acid into a conical flask; add a few drops of phenolphthalein indicator.
3. Fill a burette with 0.1 mol dm⁻³ sodium hydroxide.
4. Titrate to the endpoint — the point where the indicator just turns pink (this is the full equivalence point, where all the acid has been neutralised).
5. Add a further 25 cm³ portion of the same acid to this flask (doubling the total acid originally present, while the moles of NaOH added stays the same). The pH is then measured.

Specimen result: pH at half-equivalence point = 4.75

Handy shortcut: since pKa = −log₁₀(Ka), you can also just say pKa = pH at the half-equivalence point, directly, without converting to [H⁺] first if only pKa is asked for.

An electrochemical (voltaic) cell is built by connecting two different half-cells — each one a strip of metal dipped into a solution of its own ions — and letting electrons flow between them through an external wire. The bigger the difference in "electron-pushing power" between the two metals, the bigger the voltage (EMF) you measure.

Equipment needed:

• Two small beakers (~75 cm³) each containing a 1.0 mol dm⁻³ solution of a metal's ions
• Strips of the corresponding metals (e.g. copper, zinc, iron, silver) as electrodes
• A high-resistance voltmeter (a digital multimeter usually has this)
• Wires with crocodile clips connecting each metal strip to the voltmeter
• A salt bridge — filter paper soaked in saturated potassium nitrate solution, connecting the two beakers

• Clean the metal strips with sandpaper first to remove any oxide layer (a build-up of oxide would give an inaccurate, unstable reading).
• Fold the metal strips over the beaker edge and secure with the crocodile clips.
• Fill each beaker roughly two-thirds full with its metal-ion solution.
• Dip a strip of filter paper into saturated KNO₃ solution, then place it bridging the two beakers, making sure both ends are well immersed.
• Connect to the voltmeter, wait for the reading to steady, and record the EMF.

The voltage you actually measure can be compared against a theoretical value, calculated from tabulated standard electrode potentials (E°) found in a data book. Each half-cell has its own E° value, which represents its tendency to be reduced (to gain electrons) relative to a standard hydrogen electrode.

Worked example — zinc and iron half-cells:

1. Look up the E° values: Fe²⁺(aq) + 2e⁻ ⇌ Fe(s), E° = −0.44 V   |   Zn²⁺(aq) + 2e⁻ ⇌ Zn(s), E° = −0.76 V
2. Identify positive/negative electrodes: −0.44 V is more positive than −0.76 V, so Fe²⁺/Fe is the positive electrode and Zn²⁺/Zn is the negative electrode.
3. Apply the formula: E°(cell) = (−0.44) − (−0.76) = +0.32 V