Chemistry • Year 11 • Module 2 • Lesson 7
Standard Solutions & Dilutions
Lock in the core vocabulary, the four properties of a primary standard, and the dilution formula c1V1 = c2V2 before tackling harder questions.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: primary standard, standard solution, volumetric flask, dilution, c1V1 = c2V2, aliquot, serial dilution, quantitative transfer, moles of solute, concentration. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | A substance of very high purity, known molar mass, and stability used to make a solution of precisely known concentration. | |
| 1.2 | A solution whose concentration is known precisely, prepared from a primary standard. | |
| 1.3 | A glass flask calibrated to contain a precise volume at a specific temperature; used to make standard solutions. | |
| 1.4 | Adding solvent to a solution to reduce its concentration while keeping the number of moles of solute constant. | |
| 1.5 | The mathematical relationship stating that moles of solute are conserved when a solution is diluted. | |
| 1.6 | A precisely measured sub-sample taken from a larger solution for analysis. | |
| 1.7 | A repeated sequence of dilutions, each reducing the concentration by the same factor. | |
| 1.8 | The procedure of ensuring all solute is transferred from one vessel to another by rinsing at least three times. | |
| 1.9 | The quantity of substance (in mol) present in a solution; it remains unchanged when a solution is diluted. | |
| 1.10 | The amount of solute per unit volume of solution, measured in mol L−1. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 When a solution is diluted by adding water, the number of moles of solute decreases. T / F
2.2 A primary standard must have a high molar mass so that a weighing error of 0.001 g represents a smaller percentage of the total mass weighed. T / F
2.3 NaOH is an excellent primary standard because it is very soluble in water. T / F
2.4 In the dilution formula c1V1 = c2V2, V2 is the total final volume of the diluted solution. T / F
2.5 Both V1 and V2 must be expressed in litres when using c1V1 = c2V2. T / F
2.6 A volumetric flask is more precise than a measuring cylinder for making up a solution to an exact volume. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
conserved · dropper pipette · high purity · hygroscopic · moles · primary standard · quantitative · volumetric flask
A ___________ is a substance used to prepare a solution of precisely known concentration. It must be of ___________ so that impurities do not alter the number of moles present. Substances that absorb water vapour from the air are called ___________ and cannot be used as primary standards. To prepare a standard solution, the solid is dissolved in a beaker and then transferred ___________ to a ___________ by rinsing the beaker at least three times. The solution is made up to the calibration mark using a ___________ for the final drops. When the solution is later diluted, the number of ___________ of solute remains ___________ — only the volume changes.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 Why must a primary standard have a high molar mass?
4.2 Why is distilled water used (rather than tap water) when preparing a standard solution?
4.3 What is the underlying reason that the equation c1V1 = c2V2 holds true for a dilution?
4.4 Why would BaSO4 fail as a primary standard even though it is stable in air and has a high molar mass?
5. Complete the primary standard properties table
Fill in the missing cells. For each property, state why it is required and give one example of a substance that fails that property. 12 marks (1 per cell)
| Property | Why it matters | Example of a substance that fails & why |
|---|---|---|
| High purity | ||
| Stable in air (not hygroscopic or reactive) | ||
| High molar mass | ||
| Readily soluble in water |
6. Order the steps — preparing a standard solution
The four steps below are scrambled. Write the correct order (1–4) in the box, then fill in the missing equipment for each step. 8 marks (1 order + 1 equipment each)
| Order | Step description | Key piece of equipment |
|---|---|---|
| Add distilled water carefully until the meniscus sits exactly on the calibration mark. Use the precision tool for the last few drops. Stopper and invert to mix. | ||
| Weigh the required mass of the primary standard into a clean dry container. Record the exact mass actually obtained. | ||
| Pour the solution from the beaker into the volumetric flask. Rinse the beaker and glass rod 2–3 times with distilled water, adding each rinse to the flask. | ||
| Add a small volume (~50 mL) of distilled water to the solid in the beaker and stir until completely dissolved. |
Q1 — Term–definition match
1.1 primary standard • 1.2 standard solution • 1.3 volumetric flask • 1.4 dilution • 1.5 c1V1 = c2V2 • 1.6 aliquot • 1.7 serial dilution • 1.8 quantitative transfer • 1.9 moles of solute • 1.10 concentration.
Q2 — True / false with correction
2.1 False. Diluting a solution does NOT decrease the number of moles of solute. Adding water increases the volume but adds no solute; the moles remain constant and concentration decreases.
2.2 True. A higher molar mass means more grams per mole are weighed, so a 0.001 g balance error is a smaller percentage of the total mass — reducing the percentage error in the calculation.
2.3 False. NaOH is NOT a suitable primary standard. Although it is soluble, it fails the stability criterion because it absorbs both CO2 and water vapour from the air, continuously changing its composition and mass.
2.4 True. V2 is always the total final volume — not the volume of water added. Students must add V1 + volume of water added to find V2 when water is described as being “added.”
2.5 False. V1 and V2 do NOT need to be in litres — they must be in the same units (both L or both mL), because they appear on opposite sides of the equation and the units cancel.
2.6 True. A volumetric flask has a narrow neck that allows the meniscus to be read very precisely (±0.1 mL or better), whereas a measuring cylinder is only accurate to ±1–2 mL.
Q3 — Cloze paragraph
In order: primary standard / high purity / hygroscopic / quantitative / volumetric flask / dropper pipette / moles / conserved.
Q4.1 — Why high molar mass?
A high molar mass means a larger number of grams are needed per mole. A fixed balance error (e.g. 0.001 g) therefore represents a much smaller percentage of the total mass weighed, reducing the percentage error propagated into the concentration calculation.
Q4.2 — Why distilled water?
Distilled water is free from ionic and dissolved impurities. Using tap water would introduce ions (such as Ca2+, Mg2+, Cl−) that could react with the primary standard or alter the ionic environment, changing the actual concentration of the solution from the calculated value.
Q4.3 — Why does c1V1 = c2V2 hold?
Dilution adds only water (solvent) — no solute is added or removed. Therefore the number of moles of solute is the same before and after: n1 = n2. Since n = cV, this gives c1V1 = c2V2.
Q4.4 — Why does BaSO4 fail?
BaSO4 is essentially insoluble in water (Ksp ≈ 1.1 × 10−10). A primary standard must dissolve completely to form a homogeneous solution. An undissolved residue means the concentration of the solution is undefined — the solute in solution does not correspond to the mass weighed out.
Q5 — Four-properties table
High purity: Impurities mean the actual moles of the primary standard present are less than calculated from the total mass — the concentration will be systematically wrong from the start. Failure example: NaOH (not pure NaOH — contains Na2CO3 from CO2 absorption).
Stable in air: If the substance reacts with O2, CO2, or H2O, its composition changes between weighing and dissolving; the mass weighed no longer represents the same number of moles. Failure example: Na reacts violently with moisture; NaOH absorbs CO2 and water vapour.
High molar mass: A larger molar mass means more grams per mole are weighed, reducing the percentage error from a fixed balance uncertainty. Failure example: HCl (MM = 36.46 g mol−1) — tiny masses required, so the 0.001 g balance error is a large percentage.
Readily soluble: The substance must dissolve completely to form a homogeneous solution; an undissolved residue means the concentration in solution is undefined. Failure example: BaSO4 is practically insoluble in water.
Q6 — Correct step order and equipment
Step 1 (Weigh): Analytical balance and weighing boat. Step 2 (Dissolve): Beaker and glass rod. Step 3 (Transfer quantitatively): Volumetric flask and wash bottle. Step 4 (Make up to mark): Dropper pipette; stopper.
Scrambled order in question: Row 1 = Step 4; Row 2 = Step 1; Row 3 = Step 3; Row 4 = Step 2.