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Physical Chemistry Core Practicals

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  Edexcel IAL Chemistry

Physical Chemistry Core Practicals

🎯 The Big Idea: These four practicals are all about measuring things precisely (gas volume, temperature, or titre) and then using the maths of moles — n = mass/M, q = mcΔT, and concentration = moles/volume — to work backwards to an unknown: an unknown metal, an unknown enthalpy change, or an unknown concentration.

📋 Summary — What This Chapter Covers

  • Core Practical 1 — Molar Volume of a Gas: React a solid (e.g. a metal carbonate) with excess acid, collect the CO₂ gas produced, and use the mass–volume data to calculate the molar volume of the gas or identify an unknown metal.
  • Core Practical 2 — Enthalpy Change of Reaction: Use a simple polystyrene-cup calorimeter to measure a temperature change, then calculate the heat released/absorbed and the molar enthalpy change, correcting for heat loss with a temperature-correction graph.
  • Core Practical 3 — Determining Concentrations (Titration): Use a burette and pipette to titrate an acid against a base of known concentration, recording concordant titres, then calculate the unknown concentration using mole ratios.
  • Core Practical 4 — Preparing a Standard Solution: Accurately weigh a solid, dissolve it, and make it up to a precise volume in a volumetric flask to create a solution of known concentration (a "standard solution") — the starting point for any titration.
  • All four practicals share the same underlying skill: convert a measurement (mass, volume, temperature, titre) into moles, then use mole ratios from a balanced equation to find what you actually want.

1️⃣ Core Practical 1: Molar Volume of a Gas

The Setup — Two Ways to Collect Gas

Imagine you're reacting a metal carbonate with acid and CO₂ gas bubbles off. You need to trap and measure that gas somehow. There are two classic methods:

  • Gas syringe method: the flask is sealed with a bung connected directly to a gas syringe. As gas is produced, it pushes the plunger out, and you read the volume straight off the syringe scale.
  • Displacement of water method: gas is piped through a delivery tube into an upturned, water-filled measuring cylinder standing in a trough of water. As gas enters, it pushes water out of the cylinder, and the volume of water displaced equals the volume of gas collected.

Think of it like blowing into a plastic bag versus blowing bubbles underwater into an upside-down glass — both trap the same gas, just with different apparatus.

The Method (Sodium Carbonate + HCl Example)

The reaction used as the standard example is:

Reaction Equation
Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)
  1. Measure a fixed volume of HCl (e.g. 25.0 cm³) into a conical flask.
  2. Add a known, small mass of sodium carbonate (e.g. 0.05 g) to the flask.
  3. Immediately connect the gas syringe delivery tube (speed matters — you don't want gas escaping before you seal it!).
  4. Let the reaction go to completion.
  5. Record the volume of CO₂ produced.
  6. Repeat with increasing masses of sodium carbonate (0.10 g, 0.15 g … up to 0.50 g) to build a dataset.
⚠️ Assumptions Made in This Experiment
Every calculation here rests on three assumptions: (1) gas lost between adding the solid and connecting the tube is negligible, (2) the setup is completely airtight so no gas escapes afterwards, and (3) the reaction goes to completion. If an exam question challenges the accuracy of the method, these three are exactly what they're testing you on.

Turning Results Into a Molar Volume

You plot mass of sodium carbonate (x-axis) against volume of CO₂ produced (y-axis), ignore anomalies, and draw a line or curve of best fit. Then you pick a sensible point on the line — say, 0.35 g gives 79.0 cm³ — and work through the maths:

Step-by-step logic

① Moles of Na₂CO₃ = mass ÷ molar mass = 0.35 ÷ 106.0 = 0.0033 mol

② From the equation, 1 mol Na₂CO₃ → 1 mol CO₂, so moles of CO₂ = 0.0033 mol too

③ Convert volume: 79.0 cm³ ÷ 1000 = 0.079 dm³

Molar gas volume = volume ÷ moles = 0.079 ÷ 0.0033 = 23.93 dm³
In plain English: "for every mole of gas, how much space does it take up?"

Application: Finding an Unknown Metal

This is the classic exam twist: instead of finding the molar gas volume, you're given it (usually 24 dm³ at room temperature and pressure) and asked to identify an unknown Group 2 metal carbonate, MCO₃.

Worked Example

At room temperature and pressure, 0.950 g of a Group 2 metal carbonate, MCO₃, reacted with hydrochloric acid to produce 226.0 cm³ of carbon dioxide. Deduce the identity of the metal M.

Step 1 — Moles of CO₂ = volume (dm³) ÷ molar gas volume (dm³) = 0.226 ÷ 24 = 0.009417 mol

Step 2 — Since 1 mol MCO₃ releases 1 mol CO₂: moles of MCO₃ = 0.009417 mol

Step 3 — Molar mass of MCO₃ = mass ÷ moles = 0.950 ÷ 0.009417 = 100.9 g mol⁻¹

Step 4 — Subtract the mass of the carbonate ion (CO₃ = 12.0 + 3×16.0 = 60.0): M = 100.9 − 60.0 = 40.9 g mol⁻¹ → closest to calcium (40.1) → M = calcium

🚨 Common Trap
Students often forget Step 4 entirely and give the answer as "100.9 g/mol = the metal" — but that 100.9 is the mass of the whole carbonate, not just the metal. You MUST subtract the CO₃ part (60.0) before matching to the periodic table!
Practice Question 1
1.00 g of an unknown Group 2 metal carbonate reacts completely with excess hydrochloric acid to produce 240 cm³ of CO₂ at room temperature and pressure (molar gas volume = 24 dm³ mol⁻¹). Identify the metal.
Practice Question 2
Why must the sodium carbonate be added to the acid quickly, and the gas syringe connected immediately, in the molar volume experiment?

2️⃣ Core Practical 2: Determining Enthalpy Change of Reaction

What Is Calorimetry?

Calorimetry is just a fancy word for "measuring heat changes." A calorimeter can be as simple as a polystyrene cup (great insulator, cheap, easy to use) or as sophisticated as a vacuum flask or metal can. The polystyrene cup is the classic school-lab version because polystyrene barely conducts heat, so very little energy escapes to the surroundings while you're taking readings.

The Key Equation — Heat Transferred
q = m × c × ΔT

q = heat transferred (J)  |  m = mass of water/solution (g)  |  c = specific heat capacity (J g⁻¹ K⁻¹, water = 4.18)  |  ΔT = temperature change (K)

Plain English: "How much energy did it take to change the temperature of this much liquid by this many degrees?"

Sample Method — Zinc + Copper Sulfate Displacement

  1. Pipette 25 cm³ of 1.0 mol dm⁻³ copper(II) sulfate solution into the polystyrene cup.
  2. Weigh out roughly 6 g of zinc powder — it's in excess, so an exact mass isn't crucial.
  3. Record the initial temperature, then temperature every half-minute for 2.5 minutes (this establishes a stable "before" baseline).
  4. At exactly 3 minutes, tip in the zinc powder (don't record a reading at that exact moment — you're mid-action).
  5. Keep stirring and recording temperature every half-minute for a further 6 minutes.
🧩 Assumptions Behind the Calculation
To make the maths work, we assume: the solution behaves like pure water (specific heat capacity = 4.18 J g⁻¹K⁻¹, density = 1 g cm⁻³), the cup itself absorbs no heat, the reaction goes to completion, and there are no heat losses to the surroundings. None of these are perfectly true — that's why we need the correction graph below.

Temperature Correction Graphs — Why We Need Them

Here's the problem: many reactions aren't instant. While the reaction is still happening, the mixture is also losing heat to the surroundings the whole time. So by the time the reaction actually finishes, the measured "peak" temperature is already lower than the true maximum would have been if no heat had escaped. It's like trying to catch the exact top of a bouncing ball's trajectory while gravity is already pulling it down — you need to work backward.

The fix: plot temperature vs. time. You get a flat baseline before the reactant is added, then a rising/falling section, then a steady cooling section that slopes gently downward as heat leaks out. You draw a best-fit line through that cooling section and extrapolate it backwards until it crosses the exact time the second reactant was added. That intersection point gives you the "true" peak temperature (T₂), as if no heat had been lost at all.

Then Finish With
ΔH = q ÷ n

q = energy transferred (in kJ)  |  n = moles of the limiting reagent

This gives you the enthalpy change per mole — the number chemists actually care about.
💡 Sign Check
If temperature rose during the reaction, the reaction released heat — it's exothermic, so ΔH must be negative. If temperature fell, it's endothermic and ΔH is positive. Forgetting the negative sign for an exothermic reaction is one of the most common lost marks in this topic!
Practice Question 1
25 cm³ of 1.0 mol dm⁻³ copper(II) sulfate (excess zinc used) shows a temperature rise (corrected using the extrapolation method) of ΔT = 8.5 °C. Calculate the heat released, q, in kJ.
Practice Question 2
Explain why we need to extrapolate the cooling section of the temperature–time graph back to the moment the reactant was added, rather than just reading off the highest recorded temperature.

3️⃣ Core Practical 3: Determining Concentrations (Titration)

The Apparatus

A titration is essentially a very precise "add-a-little-at-a-time" experiment. You know the exact volume and concentration of one solution (in the conical flask, measured by volumetric pipette), and you slowly add a second solution (from the burette) until an indicator shows the reaction has just finished — the end point (equivalence point).

📏 Precision Matters
Burettes are marked to 0.10 cm³ divisions. Since they're analogue instruments, the reading uncertainty is half the smallest division: ±0.05 cm³. Because you take two readings (initial and final) to calculate the titre, the uncertainties add — giving a combined titre uncertainty of ±0.10 cm³. This "doubling of uncertainty when subtracting two measurements" is a classic idea in propagation of uncertainty, and examiners love testing it.

Step-by-Step Method

  1. Pipette a known volume (usually 20 or 25 cm³) of one solution into a conical flask.
  2. Fill the burette with the other solution, usually starting at 0.00 cm³.
  3. Add a few drops of indicator to the flask.
  4. Open the burette tap and add solution portion by portion, swirling constantly.
  5. As you approach the end point (the colour starts to change more persistently), slow right down and add dropwise — ideally you can stop after just one drop causes the permanent colour change.
  6. Repeat until you get concordant results — titres within 0.1 cm³ of each other.

Recording Results Properly

All burette readings are recorded to 2 decimal places (e.g. 23.15 not 23.1), and — because burettes are marked in 0.10 cm³ intervals — the second decimal digit should always be a 0 or a 5 (like 23.15 or 23.10, never 23.13). The first "rough" titration is done quickly to find the approximate end point, then discarded from the final average since it's usually too high (you weren't being precise near the end point yet).

Worked Example

25.0 cm³ of hydrochloric acid was titrated with a 0.200 mol dm⁻³ solution of sodium hydrogencarbonate, NaHCO₃.

NaHCO₃ + HCl → NaCl + H₂O + CO₂

Concordant titres from Run 2 and Run 3 averaged to 22.80 cm³. Calculate the concentration of the acid.

Step 1 — Average titre = (22.80 + 22.80) ÷ 2 = 22.80 cm³

Step 2 — Moles NaHCO₃ = (22.80 ÷ 1000) × 0.200 = 4.56 × 10⁻³ mol

Step 3 — Ratio NaHCO₃ : HCl is 1:1, so moles HCl = 4.56 × 10⁻³ mol too

Step 4 — Concentration = moles ÷ volume = 4.56×10⁻³ ÷ (25.0/1000) = 0.182 mol dm⁻³

⚠️ Watch the Stoichiometry Ratio!
Step 3 above only works because the ratio is 1:1. If the balanced equation had a 2:1 or 1:2 ratio, you'd need to scale the moles up or down accordingly before dividing by volume. Always check the balanced equation before assuming the moles are equal!
Practice Question 1
A student titrates 25.0 cm³ of NaOH solution against 0.100 mol dm⁻³ HCl. The concordant average titre is 18.40 cm³. HCl + NaOH → NaCl + H₂O. Calculate the concentration of the NaOH.
Practice Question 2
A titration gives initial burette reading 0.20 cm³ and final reading 23.00 cm³. State the titre with correct uncertainty, and explain why the uncertainty is ±0.10 cm³ rather than ±0.05 cm³.

4️⃣ Core Practical 4: Preparing a Standard Solution

What Is Volumetric Analysis?

Volumetric analysis uses the volume and concentration of one solution (a "volumetric" or "standard" solution) to find the concentration of an unknown one. Before you can titrate anything, though, you first need to make that known solution — precisely and accurately. That's what this practical is about.

The 5-Step Method

  1. Weigh a precise mass of the solid on a 3-decimal-place balance.
  2. Dissolve it in a small volume of water in a beaker, stirring with a glass rod until fully dissolved.
  3. Transfer the solution to a volumetric flask using a funnel — don't lose any solid on the way!
  4. Rinse the beaker and glass rod with distilled water, adding all the rinsings to the flask (this ensures no solute is left behind — every last bit must end up in the flask).
  5. Make up to the mark: add more water carefully until the bottom of the meniscus sits exactly on the graduation (scratch) mark, then stopper and invert/mix thoroughly.
👁️ Reading the Meniscus
When making up to the mark, you must view the graduation line at eye level and line up the bottom of the meniscus (the curved liquid surface) with the mark — not the top. Looking at it from above or below gives a parallax error that throws off your whole concentration calculation.

The Core Concentration Formula

Concentration
Concentration (mol dm⁻³) = moles ÷ volume (dm³)
The amount of solute (in moles) packed into every 1 dm³ (1 litre) of solution. A concentrated solution has a lot of solute per dm³; a dilute one has little.
Worked Example

Calculate the mass of sodium hydroxide, NaOH, required to prepare 250 cm³ of a 0.200 mol dm⁻³ solution.

Step 1 — Moles needed = concentration × volume = 0.200 mol dm⁻³ × 0.250 dm³ = 0.0500 mol

Step 2 — Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g mol⁻¹

Step 3 — Mass = moles × molar mass = 0.0500 × 40.00 = 2.00 g

🔁 The Three Ways to Express Concentration
Concentration can be given as: moles per unit volume (mol dm⁻³ — most common in this chapter), mass per unit volume (g dm⁻³), or parts per million (ppm — used for very dilute solutions, like pollutant levels). Always check which unit a question wants before you answer!
Practice Question 1
Calculate the mass of anhydrous sodium carbonate, Na₂CO₃ (Mr = 106.0), required to make 500 cm³ of a 0.150 mol dm⁻³ standard solution.
Practice Question 2
Explain why the beaker used to dissolve the solid must be rinsed, with the rinsings added to the volumetric flask, rather than just tipping the solution across and stopping there.

🧠 What to Memorise

Term / FormulaMeaning
q = mcΔTHeat transferred = mass × specific heat capacity × temperature change
ΔH = q ÷ nEnthalpy change per mole = heat transferred ÷ moles of limiting reagent
c (water) = 4.18 J g⁻¹ K⁻¹Specific heat capacity of water/dilute aqueous solutions (assumed)
Moles = mass ÷ MrConverts a measured mass into moles
Concentration = moles ÷ volume (dm³)mol dm⁻³ — the standard unit of solution concentration
Molar gas volume = volume ÷ molesVolume occupied by one mole of gas (≈24 dm³ at RTP)
CalorimetryTechnique for measuring enthalpy (heat) changes in reactions
Concordant resultsTitres within 0.1 cm³ of each other — used to calculate the average
End point / equivalence pointThe moment the reaction is exactly complete, shown by indicator colour change
Standard/volumetric solutionA solution whose concentration is known precisely
Burette uncertainty±0.05 cm³ per reading → ±0.10 cm³ per titre (two readings subtracted)
Limiting reagentThe reactant that runs out first and controls how much product forms — used as "n" in ΔH = q/n
Gas syringe Displacement of water Polystyrene cup calorimeter Temperature correction graph Volumetric pipette Burette Volumetric flask Meniscus

✅ Concepts Checklist

🎯 Exam Tips & Common Mistakes

⚠️ Forgetting to subtract the carbonate ion mass
When finding an unknown metal M in MCO₃, examiners deliberately choose values where the whole carbonate's molar mass looks tantalisingly close to a real metal's atomic mass. Always subtract CO₃ (60.0 g mol⁻¹) before matching to the periodic table.
⚠️ Missing the negative sign for exothermic reactions
A temperature rise means the reaction released energy — ΔH must be negative. This is one of the most frequently dropped marks in enthalpy questions, even when every number is otherwise correct.
⚠️ Reading the wrong part of the meniscus
Always read the bottom of the meniscus at eye level for colourless/pale solutions (for very dark solutions like potassium manganate(VII), the top is used instead — but the standard rule taught here is bottom of the meniscus).
✅ What examiners want to see in method descriptions
Specific volumes/masses (not "some acid"), correct apparatus names (volumetric pipette, not "measuring cylinder" for precise volumes), and an explicit mention of repeating until concordant results are achieved.
⚠️ Uncertainty doubling
Any time a value is calculated by subtracting two measured readings (like a titre, or an initial/final burette reading), the uncertainties add together — don't just quote the single-reading uncertainty.
⚠️ Discarding the "rough" titration incorrectly
The rough titre is used to find an approximate end point quickly — it's normal practice to discard it from the final average because it's collected less carefully and is often too high (over-shot past the true end point).
⚠️ Forgetting to identify the limiting reagent
In ΔH = q/n, "n" must be moles of the limiting reagent, not just whichever reactant is mentioned first. If one reactant is described as being "in excess," the other one is automatically the limiting reagent — but always double check by comparing mole ratios.
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