A forensic toxicologist testing a blood sample can't use a vague "roughly 1 mol/L" acid — the result has to stand up in court. A pharmaceutical lab dissolving a drug standard can't afford to be off by 0.1%. This is why primary standards and volumetric flasks exist: chemistry where precision is not optional.
📚 Core Content
A standard solution is one whose concentration is known precisely. You can make a standard solution from scratch only if you start with a substance that meets strict criteria — a primary standard.
Most chemicals you'd find in a lab cannot be used as primary standards. Concentrated HCl, for example, is a liquid of variable concentration that absorbs water from air — you could never weigh out a precise amount. NaOH absorbs both CO₂ and water from air, changing its mass continuously.
A primary standard must satisfy all four of the following:
| Property | Why it matters | Example failure |
|---|---|---|
| High purity | If the substance contains impurities, the actual moles present will differ from the calculated value — the concentration will be wrong from the start. | NaOH absorbs CO₂ — not pure NaOH |
| Stable (doesn't react with air) | If the substance reacts with O₂, CO₂, or H₂O in the atmosphere, its composition changes between weighing and dissolving. The mass you weighed no longer represents the same number of moles. | Na reacts violently with moisture |
| High molar mass | A higher molar mass means you need to weigh out more grams per mole. This reduces the percentage error from the balance — a 0.001 g weighing error matters far less when you're weighing 200 g than when you're weighing 4 g. | HCl (MM = 36.46) — tiny masses, huge % error |
| Readily soluble in water | The substance must dissolve completely and quickly to form a homogeneous solution. An undissolved residue means not all the solute is in solution — the concentration is undefined. | BaSO₄ is insoluble — can't make a solution |
This procedure is examinable. You may be asked to describe it and justify each step.
When you add water to a solution, you increase its volume but the number of moles of solute stays the same. This is the key insight behind dilution. The concentration must decrease to keep the moles constant.
A serial dilution is a repeated sequence of dilutions, each reducing the concentration by the same factor. It's used to produce a range of very low concentrations from a concentrated stock — for example, when preparing calibration standards for spectroscopy.
🧮 Worked Examples
🧪 Activities
1 50.0 mL of 3.00 mol L⁻¹ NaOH is diluted to 300 mL. Calculate the final concentration.
2 What volume of 12.0 mol L⁻¹ HCl must be diluted to 1.00 L to give 0.250 mol L⁻¹ HCl?
3 25.0 mL of a 0.400 mol L⁻¹ solution is diluted by adding 75.0 mL of water. What is the new concentration?
4 A student dilutes 10.0 mL of stock solution to 500 mL and measures the final concentration as 0.0200 mol L⁻¹. What was the concentration of the original stock solution?
Type your working for all four problems:
Complete in your workbook.
Write your evaluation for A, B, C, and D below:
Answer in your workbook.
❓ Multiple Choice
1. Which property of a primary standard reduces the percentage error in weighing?
2. Why is NaOH unsuitable as a primary standard?
3. 200 mL of 1.50 mol L⁻¹ NaCl is diluted to 600 mL. What is the new concentration?
4. Which statement about the dilution formula c₁V₁ = c₂V₂ is correct?
5. A student needs 250 mL of 0.100 mol L⁻¹ HCl, starting from 2.00 mol L⁻¹ HCl. What volume of the concentrated acid is required?
✍️ Short Answer
6. A student prepares a standard solution of potassium iodate (KIO₃) by dissolving 1.070 g of KIO₃ in distilled water and making up to 250.0 mL in a volumetric flask. (a) Calculate the concentration of the standard solution. (b) State two properties of KIO₃ that make it suitable as a primary standard. (K = 39.098, I = 126.90, O = 15.999) 4 MARKS
Type your answer:
Answer in workbook.
7. A laboratory technician needs to prepare 500 mL of 0.0500 mol L⁻¹ sulfuric acid from a concentrated stock solution of 9.80 mol L⁻¹. (a) Calculate the volume of concentrated acid needed. (b) Describe the procedure the technician should follow, identifying the key piece of equipment used to ensure accuracy. 5 MARKS
Type your answer:
Answer in workbook.
8. A student performs a 1:5 serial dilution of a 1.00 mol L⁻¹ glucose solution — that is, at each step they take 10.0 mL of the current solution and make it up to 50.0 mL. (a) Calculate the concentration after the first dilution. (b) Calculate the concentration after the third dilution. (c) How many moles of glucose are present in 25.0 mL of the third dilution? 4 MARKS
Type your answer:
Answer in workbook.
A. NaOH — Fails: stability in air. NaOH absorbs both CO₂ (reacting to form Na₂CO₃) and water vapour from the atmosphere. The mass you weigh out is therefore not pure NaOH — it contains Na₂CO₃ and adsorbed water. The moles of NaOH present are unknown, so an accurate concentration cannot be calculated.
B. KHP — Passes all four criteria. It is available in high purity, is stable at room temperature in dry air, has a high molar mass (204.22 g mol⁻¹) that minimises weighing error, and is freely soluble in water. KHP is one of the most widely used primary standards in titrimetric analysis.
C. HCl — Fails: purity and stability. Concentrated HCl is a solution of HCl gas in water — its concentration varies with temperature, age, and storage conditions. You cannot weigh out a precise number of moles. It is also volatile, losing HCl gas when opened. Fails both purity and stability criteria.
D. BaSO₄ — Fails: solubility. Although BaSO₄ meets the first three criteria (pure, stable, high MM = 233.39 g mol⁻¹), it is essentially insoluble in water (Ksp ≈ 1.1 × 10⁻¹⁰). You cannot make a homogeneous solution of a known concentration from an insoluble substance.
1. C — High molar mass means more grams per mole — a weighing error of 0.001 g is a smaller percentage of 204 g than of 40 g.
2. B — NaOH reacts with atmospheric CO₂ and absorbs water vapour, changing its composition and mass over time.
3. D — c₂ = (1.50 × 200) ÷ 600 = 300 ÷ 600 = 0.500 mol L⁻¹.
4. A — Adding water increases volume but adds no solute — moles are conserved, so n₁ = n₂, giving c₁V₁ = c₂V₂.
5. C — V₁ = (0.100 × 250) ÷ 2.00 = 25.0 ÷ 2.00 = 12.5 mL.
Q6 (4 marks):
(a) MM(KIO₃) = 39.098 + 126.90 + 3(15.999) = 39.098 + 126.90 + 47.997 = 213.995 g mol⁻¹ n = m ÷ MM = 1.070 ÷ 213.995 = 0.005000 mol V = 0.2500 L; c = 0.005000 ÷ 0.2500 = 0.02000 mol L⁻¹(b) Any two of: high purity (available in reagent-grade form); stable in air (does not absorb CO₂ or water); high molar mass (214 g mol⁻¹ — minimises % weighing error); readily soluble in water.
Q7 (5 marks):
(a) V₁ = c₂V₂ ÷ c₁ = (0.0500 × 500) ÷ 9.80 = 25.0 ÷ 9.80 = 2.55 mL(b) Using a pipette (or measuring cylinder), measure out 2.55 mL of concentrated H₂SO₄. Add this slowly to approximately 400 mL of distilled water in a large beaker — always add acid to water. Stir to mix and allow to cool. Transfer to a 500 mL volumetric flask, rinsing the beaker 3 times with distilled water. Make up to the calibration mark with distilled water using a dropper pipette. Stopper and invert to mix. The key piece of equipment is the volumetric flask, which ensures the volume is accurate to ±0.1 mL.
Q8 (4 marks):
(a) c₂ = (1.00 × 10.0) ÷ 50.0 = 0.200 mol L⁻¹ (b) After step 2: (0.200 × 10.0) ÷ 50.0 = 0.0400 mol L⁻¹ After step 3: (0.0400 × 10.0) ÷ 50.0 = 8.00 × 10⁻³ mol L⁻¹ (c) n = c × V = 8.00 × 10⁻³ × 0.0250 = 2.00 × 10⁻⁴ molTick when you've finished all activities and checked your answers.