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Module 2 · L7 of 20 ~35 min ⚡ +50 XP in Learn · +25 to complete

Standard Solutions & Dilutions

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.

Today's hook, 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. Precision matters.

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Warm up and think first

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Recall, your gut answer first

+5 XP warm-up

When you make a cup of cordial, you start with a concentrated syrup and add water to dilute it. If you have 50 mL of cordial syrup that is 10× strength, and you want to make a normal-strength drink, how much water would you need to add? What principle governs the relationship between the concentrated and diluted versions?

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The formulas, and the words for them

Formula reference · this lesson

core formulas

From L06, still needed

$$c = n \div V$$

c = concentration (mol L⁻¹)

n = moles of solute (mol)

V = volume of solution (L)

New this lesson, dilution

$$c_1 V_1 = c_2 V_2$$

c₁ = initial concentration (mol L⁻¹)

V₁ = initial volume (L or mL)

c₂ = final concentration (mol L⁻¹)

V₂ = final volume (L or mL)

⚠️ V must be in litres for c = n ÷ V. For c₁V₁ = c₂V₂ the units must match on both sides (both L or both mL).

What you'll master

Know

Key facts

Definition of a primary standard
Four properties a primary standard must have
Examples: Na₂CO₃, KIO₃, oxalic acid
The dilution formula c₁V₁ = c₂V₂

Understand

Concepts

Why moles of solute are conserved during dilution
Why most reagents can't be used as primary standards
The role of a volumetric flask in accurate preparation

Can do

Skills

Describe the preparation of a standard solution
Identify whether a substance qualifies as a primary standard
Use c₁V₁ = c₂V₂ to solve dilution problems

Key terms

Primary standard

A substance of very high purity, known molar mass, and stability used to make a solution of precisely known concentration.

Standard solution

A solution whose concentration is known precisely, prepared from a primary standard.

Volumetric flask

A glass flask calibrated to contain a precise volume at a specific temperature; used to make standard solutions.

Dilution

Adding solvent to a solution to reduce its concentration; moles of solute remain constant.

c₁V₁ = c₂V₂

Dilution equation: moles before dilution = moles after dilution, so c₁V₁ = c₂V₂.

Aliquot

A precisely measured sub-sample taken from a larger solution for analysis.

Primary standards, all four criteria

Primary Standards

core concept

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.

The Four Properties of a Primary Standard

A primary standard must satisfy all four of the following:

✓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. Example failure: 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. Example failure: 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. Example failure: 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. Example failure: BaSO₄ is insoluble, can't make a solution.

Common primary standards in NSW HSC

Sodium carbonate (Na₂CO₃, MM = 105.99), potassium iodate (KIO₃, MM = 214.00), oxalic acid dihydrate (H₂C₂O₄·2H₂O, MM = 126.07), potassium hydrogen phthalate (KHP, MM = 204.22). All have high molar masses and are stable solids at room temperature.

Preparing a Standard Solution, Step by Step

This procedure is examinable. You may be asked to describe it and justify each step.

Weigh the primary standard accurately use an analytical balance to weigh the required mass into a clean, dry weighing boat. Record the exact mass, use this value in calculations, not the theoretical target mass.
Dissolve in a small volume of distilled water transfer the solid to a beaker. Add ~50 mL of distilled water and stir until completely dissolved. Distilled water prevents ionic impurities.
Transfer quantitatively to a volumetric flask pour the solution into a volumetric flask of the desired volume. Rinse the beaker and glass rod 2–3 times with distilled water, adding each rinse to the flask. This ensures all solute is transferred.
Make up to the mark with distilled water add distilled water carefully until the bottom of the meniscus sits exactly on the calibration mark. Use a dropper pipette for the final additions. Stopper and invert several times to mix thoroughly.

Why a volumetric flask, not a measuring cylinder?

Volumetric flasks are calibrated to contain a precise volume at a specific temperature (usually 20°C). The long, narrow neck means the meniscus can be read very precisely. A measuring cylinder is accurate to ±1–2 mL; a volumetric flask is accurate to ±0.1 mL or better.

A primary standard must satisfy all four criteria: high purity, stable in air, high molar mass, and readily soluble in water. Common examples include Na₂CO₃, KIO₃, KHP and oxalic acid. NaOH and HCl cannot be primary standards because they absorb CO₂/H₂O or have variable concentration.

Pause, copy the highlighted definition into your book before moving on.

Quick check: Which statement best explains why NaOH cannot be used as a primary standard?

Did you get this? True or false: a measuring cylinder is just as accurate as a volumetric flask for preparing standard solutions.

Odd one out three of these substances meet all the criteria for a primary standard. Which one does not?

Dilution, and why the moles do not change

Dilution, c₁V₁ = c₂V₂

core concept

We just saw how to prepare a standard solution from a primary standard using precise mass and a volumetric flask. That raises a question: if you already have a concentrated standard solution, how do you produce a more dilute version without re-weighing? This card answers it → with the dilution equation c₁V₁ = c₂V₂.

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.

Derivation

Since moles are conserved: n₁ = n₂ | c₁V₁ = c₂V₂ (because n = cV for both). Important: V₁ and V₂ must be in the same units (both L or both mL), but they don't have to be litres, because they appear on both sides and the units cancel.

Serial Dilution

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.

Example, 1:10 serial dilution

Starting at 1.00 mol L⁻¹:
Step 1: 10 mL of stock → 100 mL → 0.100 mol L⁻¹
Step 2: 10 mL of Step 1 → 100 mL → 0.0100 mol L⁻¹
Step 3: 10 mL of Step 2 → 100 mL → 0.00100 mol L⁻¹
Each step multiplies the concentration by ×0.1 (or ÷10).

Dilution equation: c₁V₁ = c₂V₂ (moles of solute are conserved). V₂ is the total final volume (not just water added). V₁ and V₂ must be in the same units; c₂ must always be less than c₁ after dilution.

Pause, write the highlighted equation into your book.

Fill the blanks: drag each token into the matching blank.

moles of solute V₂ total final volume less than c₁

In dilution, the ___ stay constant, only volume changes. So c₁V₁ = c₂___. V₂ is the ___, and after dilution c₂ must be ___.

Did you get this? True or false: if a question says "100 mL of stock is diluted by adding 400 mL of water", V₂ should be 400 mL.

Predict, then reveal

Predict, then reveal+8 XP

1 · Predict

2 · Reveal

3 · Compare

A student takes 25.0 mL of 1.20 mol L⁻¹ HCl and dilutes it to a final volume of 250.0 mL. Predict the new concentration in mol L⁻¹.

Confidence: 50%

Three worked examples

Worked examples · reveal as you go

Worked example 1 · preparing a standard solution +5 XP on full reveal

A student dissolves 2.650 g of anhydrous sodium carbonate (Na₂CO₃) in distilled water and makes up to 250.0 mL in a volumetric flask. Calculate the exact concentration of the resulting standard solution. (Na = 22.990, C = 12.011, O = 15.999)

$\mathrm{MM(Na_2CO_3)} = 2(22.990) + 12.011 + 3(15.999) = 105.989\ \mathrm{g\ mol^{-1}}$

Calculate molar mass

$n = m \div MM = 2.650 \div 105.989 = 0.025003\ \mathrm{mol}$

Convert mass to moles

$V = 250.0\ \mathrm{mL} \div 1000 = 0.2500\ \mathrm{L}$

Convert volume to litres

$c = n \div V = 0.025003 \div 0.2500 =$ 0.1000 mol L⁻¹ ✓

Calculate concentration

Worked example 2 · basic dilution calculation +5 XP on full reveal

100 mL of 2.00 mol L⁻¹ hydrochloric acid (HCl) is diluted to a final volume of 500 mL. Calculate the concentration of the diluted solution.

$c_1 = 2.00\ \mathrm{mol\ L^{-1}}$ | $V_1 = 100\ \mathrm{mL}$ | $V_2 = 500\ \mathrm{mL}$ | $c_2 = ?$

Identify variables, units match (both mL)

$c_1V_1 = c_2V_2$ → $c_2 = c_1V_1 \div V_2$

Rearrange formula

$c_2 = (2.00 \times 100) \div 500 = 200 \div 500$

Substitute

$c_2 =$ 0.400 mol L⁻¹ ✓

Calculate (c₂ < c₁ as expected)

Worked example 3 · finding volume needed for dilution +5 XP on full reveal

What volume of 6.00 mol L⁻¹ sulfuric acid (H₂SO₄) must be diluted to 250 mL to give a 0.500 mol L⁻¹ solution?

$c_1 = 6.00\ \mathrm{mol\ L^{-1}}$ | $V_1 = ?$ | $c_2 = 0.500\ \mathrm{mol\ L^{-1}}$ | $V_2 = 250\ \mathrm{mL}$

Identify variables

$V_1 = c_2V_2 \div c_1$

Rearrange

$V_1 = (0.500 \times 250) \div 6.00 = 125 \div 6.00$

Substitute

$V_1 =$ 20.8 mL of the 6.00 mol L⁻¹ H₂SO₄ stock ✓

Calculate

Safety boundary. Diluting strong or concentrated acids is technician or teacher work, not an independent student task. Concentrated acid is always added slowly to water (never water to acid), in a fume cupboard, wearing safety glasses, gloves and a lab coat. In class you work with the pre-diluted solution your technician has prepared.

The three traps that cost marks

Common errors · the 3 traps that cost marks

Mixing up V₁ and V₂ in the dilution formula

V₁ is the volume of the concentrated solution you measure out, not the final volume. V₂ is always the larger, final volume. Students sometimes reverse these and get an answer that implies the diluted solution is more concentrated, which is impossible.

Fix: Sanity-check before substituting. After dilution, c₂ must always be smaller than c₁. If your c₂ > c₁, you've swapped V₁ and V₂.

Using V₂ = volume of water added, not final volume

If a question says "100 mL of stock is diluted by adding 400 mL of water," students sometimes put V₂ = 400 mL. The correct V₂ is the total final volume = 100 + 400 = 500 mL.

Fix: V₂ = total final volume = V₁ + volume of water added. "Made up to 500 mL" → V₂ = 500 mL. "400 mL of water added" → V₂ = V₁ + 400 mL.

Listing primary standard properties without explaining them

Short-answer questions almost always ask you to "state AND explain" each property. Writing "high purity" without explaining why purity matters scores 0–1 marks. The explanation is where the marks are.

Fix: For each property, write a two-part answer: "It must be [property] because [consequence if it isn't]."

Quick-fire drill, then revisit your thinking

Quick-fire practice · 5 reps +2 XP per reveal

50.0 mL of 3.00 mol L⁻¹ NaOH is diluted to 300 mL. Calculate the final concentration.

What volume of 12.0 mol L⁻¹ HCl must be diluted to 1.00 L to give 0.250 mol L⁻¹ HCl?

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?

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?

A 1.000 mol L⁻¹ stock solution is used to prepare 250.0 mL of 0.0500 mol L⁻¹ working solution. What volume of stock is needed?

Revisit your thinking

Earlier you were asked: For 50 mL of 10× cordial syrup, how much water makes a normal-strength drink?

The answer: add 450 mL of water (making 500 mL total). The principle is that moles of solute are conserved, c₁V₁ = c₂V₂. If c₁ = 10×, c₂ = 1×, V₁ = 50 mL, then V₂ = (10 × 50) ÷ 1 = 500 mL. The same formula applies to all chemical dilutions, and it's the foundation of the standard solution and dilution calculations in this lesson.

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✓

Practice questions

Multiple choice

+2 XP per correct · +5 bonus if perfect

Pick your answer, then rate your confidence. That tells the system what to drill next.

Short answer

UnderstandLower-order4 marks

Q1. 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)

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ApplyMid-order5 marks

Q2. A laboratory technician needs to prepare 500 mL of 0.0500 mol L⁻¹ sulfuric acid from a 9.80 mol L⁻¹ acid stock. (a) Calculate the volume of stock acid needed. (b) Describe the procedure the technician should follow (including the safety controls for handling strong acid), identifying the key piece of equipment used to ensure accuracy. Note: diluting strong acid is technician/teacher work; students use the prepared solution.

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AnalyseMid-order4 marks

Q3. 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?

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EvaluateHigher-order5 marks

Q4. A student standardises a NaOH solution using KHP (potassium hydrogen phthalate, MM = 204.22 g mol⁻¹). They dissolve 0.408 g of KHP in water and titrate it with the NaOH solution, using 20.00 mL of NaOH to reach the endpoint. They calculate the NaOH concentration as 0.100 mol L⁻¹. A second student checks their working and says: "Your calculation is wrong, you forgot to account for the 1:1 mole ratio." Evaluate the original student's calculation and the second student's criticism. (KHP reacts with NaOH in a 1:1 mole ratio)

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CreateHigher-order6 marks

Q5. A chemist needs to prepare a series of standard solutions of Na₂CO₃ at concentrations of 0.100, 0.0500, 0.0250, and 0.0125 mol L⁻¹, each in 100 mL volumes. They have a 0.200 mol L⁻¹ stock solution of Na₂CO₃ available. Design the most efficient procedure to prepare all four solutions using serial dilution, showing all calculations for each step.

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📖 Comprehensive answers (click to reveal)

Multiple choice, drill bank

MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set. From the lesson bank: high molar mass means smaller % weighing error; NaOH cannot be a primary standard (absorbs CO₂/H₂O); dilution applies c₁V₁ = c₂V₂; moles conserved during dilution; sanity-check c₂ < c₁.

Short answer model answers

Q1 (4 marks):

(a) MM(KIO₃) = 39.098 + 126.90 + 3(15.999) = 213.995 g mol⁻¹
n = 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.

Q2 (5 marks):

(a) V₁ = c₂V₂ ÷ c₁ = (0.0500 × 500) ÷ 9.80 = 2.55 mL

(b) This dilution is carried out by the technician or teacher, not independently by students. Working in a fume cupboard and wearing safety glasses, gloves and a lab coat, the technician pipettes 2.55 mL of the 9.80 mol L⁻¹ acid and adds it slowly to ~400 mL of distilled water in a large beaker (always acid to water, never water to acid, because the dilution is strongly exothermic). After stirring and cooling, the solution is transferred to a 500 mL volumetric flask, the beaker rinsed 3 times into the flask, and made up to the calibration mark with distilled water using a dropper pipette, then stoppered and inverted to mix. The key piece of equipment is the volumetric flask, which fixes the final volume accurately. Students then work with this prepared solution.

Q3 (4 marks):

(a) c₂ = (1.00 × 10.0) ÷ 50.0 = 0.200 mol L⁻¹
(b) Step 2: (0.200 × 10.0) ÷ 50.0 = 0.0400 mol L⁻¹; 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 mol

Q4 (5 marks):

n(KHP) = 0.408 ÷ 204.22 = 1.998 × 10⁻³ mol ≈ 2.00 × 10⁻³ mol [1]
Since KHP:NaOH = 1:1, n(NaOH) = 2.00 × 10⁻³ mol [1]
c(NaOH) = 2.00 × 10⁻³ ÷ 0.02000 = 0.100 mol L⁻¹ [1]

The original student's calculation is correct, the 1:1 mole ratio WAS used (n(NaOH) = n(KHP)). The second student's criticism is wrong [1]. The 1:1 ratio was implicitly applied [1].

Q5 (6 marks · Band 6):

Using serial dilution where each step halves the concentration (take 50.0 mL from each and make up to 100 mL): [1 for identifying the pattern]

Step 1: V₁ = (0.100 × 100) ÷ 0.200 = 50.0 mL of stock → make to 100 mL → 0.100 mol L⁻¹ [1]
Step 2: Take 50.0 mL of 0.100 → make to 100 mL → 0.0500 mol L⁻¹ [1]
Step 3: Take 50.0 mL of 0.0500 → make to 100 mL → 0.0250 mol L⁻¹ [1]
Step 4: Take 50.0 mL of 0.0250 → make to 100 mL → 0.0125 mol L⁻¹ [1]

Each step: use a 50 mL pipette to transfer to a 100 mL volumetric flask, add distilled water to the mark, stopper and invert. This serial halving is efficient, only one calculation type is used at each step [1].

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