A standard solution is a solution whose concentration is known exactly. It is prepared by dissolving a weighed solute and making up to the mark in a volumetric flask.
Dilution adds water without changing the moles of solute, so M1V1 = M2V2.
This page covers one Form 4 content standard: standard solution and dilution. A standard solution is the reference you titrate against, so preparing one correctly and calculating dilutions accurately are practical skills the exam tests in both Paper 2 and the practical assessment. The mathematics is short, one conservation idea and one formula, but the method steps must be precise.
What a standard solution is
A standard solution is a solution whose concentration is known accurately. You make one when you need a reliable, exactly known molarity, for example, to use in a titration. It is prepared from a solute that can be weighed precisely and dissolved to a known total volume.
The key apparatus is the volumetric flask (also called a graduated flask), which is calibrated to contain one exact volume, such as 250.0 cm³, when filled to the etched mark.
Preparing a standard solution, the method
The steps below are frequently examined as an ordered list, so learn them in sequence:
- Weigh the required mass of solute accurately using an electronic balance.
- Dissolve the solute in a small volume of distilled water in a beaker, stirring with a glass rod.
- Transfer the solution into the volumetric flask through a filter funnel; rinse the beaker and rod with distilled water and add the rinsings, so no solute is left behind.
- Add distilled water until the level is close to the mark, then use a dropper to bring the bottom of the meniscus exactly onto the graduation mark.
- Stopper and invert the flask several times to mix thoroughly, giving a solution of uniform, known concentration.
Two details earn marks: rinsing the beaker (all the solute must reach the flask) and reading the meniscus at eye level (so the volume is exact).
Dilution: why moles are conserved
Dilution means adding more water (solvent) to a solution to lower its concentration. Adding water does not add or remove any solute, so the number of moles of solute stays the same before and after dilution. Only the volume, and therefore the concentration, changes.
Because moles = molarity × volume, and the moles are unchanged:
M₁V₁ = M₂V₂
where M₁, V₁ are the molarity and volume before dilution, and M₂, V₂ are those after. As long as both volumes use the same unit, you do not even need to convert to dm³ here.
Worked example
Question. A student has 25.0 cm³ of 2.0 mol dm⁻³ hydrochloric acid. She dilutes it with distilled water to a total volume of 250 cm³ in a volumetric flask. Calculate the molarity of the diluted acid.
Step 1, Write down the known values. M₁ = 2.0 mol dm⁻³, V₁ = 25.0 cm³, V₂ = 250 cm³, M₂ = ?
Step 2, Apply the dilution formula. M₁V₁ = M₂V₂, so M₂ = (M₁V₁) ÷ V₂.
Step 3, Substitute. M₂ = (2.0 × 25.0) ÷ 250 = 50 ÷ 250 = 0.2 mol dm⁻³.
Step 4, Sense check. The volume increased tenfold (25 → 250 cm³), so the concentration should fall tenfold (2.0 → 0.2 mol dm⁻³). It does.
Answer. The molarity of the diluted acid is 0.2 mol dm⁻³.
Practice question
Question. What volume of 1.0 mol dm⁻³ sodium hydroxide must be diluted with distilled water to prepare 500 cm³ of 0.1 mol dm⁻³ sodium hydroxide solution?
Answer. Use M₁V₁ = M₂V₂ with M₁ = 1.0 mol dm⁻³, M₂ = 0.1 mol dm⁻³, V₂ = 500 cm³. Then V₁ = (M₂V₂) ÷ M₁ = (0.1 × 500) ÷ 1.0 = 50 cm³. Measure 50 cm³ of the 1.0 mol dm⁻³ solution with a pipette or burette, transfer it to a 500 cm³ volumetric flask, and add distilled water up to the mark.
Exam tip
For dilution, the moles of solute do not change, say this if a question asks you to explain the calculation, because it is the reason M₁V₁ = M₂V₂ works. When you use the formula, both volumes may stay in cm³ as long as they match, but any concentration you quote must be in mol dm⁻³. For the preparation method, always include rinsing the beaker and glass rod and reading the meniscus at eye level; these two steps are the ones most often left out and each is worth a mark.
Common variations to expect
The exam recombines the same ideas in a few predictable ways. It may give you a mass of solute and a flask volume and ask for the molarity of the standard solution, that is really the concentration calculation from standard 6.3, so find the moles (mass ÷ molar mass) and divide by the volume in dm³. It may give you a concentrated stock solution and a target and ask what volume to take, that is the dilution formula rearranged for V₁, exactly as in the practice question. Or it may show a diagram of the preparation and ask you to identify apparatus or spot a mistake, such as failing to rinse the beaker or filling past the mark.
Whichever form appears, anchor your answer on two sentences: a standard solution has an accurately known concentration made up to a fixed volume in a volumetric flask, and dilution conserves moles so M₁V₁ = M₂V₂. From those two ideas every question on this standard follows. This is also the point where careful practical technique starts to matter for marks, so rehearse the ordered method until you can write it without prompting.
Where this fits
This is content standard 6.4 of the Acids, Bases and Salts chapter. It uses the molarity from standard 6.3 and prepares the accurately known solutions you will titrate in standard 6.5. Practise the calculations in the worked examples and check the method steps against the practice questions. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, our teachers rehearse the preparation method and the dilution formula together, because Paper 2 often pairs the calculation with the practical steps in one question.
Quick recap
- Standard solution: concentration known accurately; made up to the mark in a volumetric flask.
- Method essentials: weigh, dissolve, transfer + rinse, make up to the mark at eye level, mix.
- Dilution adds water; moles of solute are conserved.
- M₁V₁ = M₂V₂, volumes may share any unit; concentrations in mol dm⁻³.
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