Preparing a standard solution

A standard solution is one whose concentration is known accurately, and preparing one is both a calculation and a Paper 3 practical skill in SPM Chemistry. Our online one-to-one teachers make sure you can do the calculation and describe the method precisely, because the exam awards marks for both the number and the correct apparatus.

When you use it

You prepare a standard solution whenever a titration or a known-concentration experiment needs an exact solution, usually from a solid solute such as anhydrous sodium carbonate. The calculation tells you the mass to weigh; the method tells you how to reach the exact final volume.

The formula

mass = moles x molar mass, where moles = molarity x volume (in dm cubed)

Putting them together:

mass (g) = molarity (mol dm⁻³) x volume (dm³) x molar mass (g mol⁻¹)

Units

Volume must be in dm cubed (convert cm cubed by dividing by 1000), molarity in mol dm⁻³, molar mass in g mol⁻¹, and the mass comes out in grams.

The practical steps

1. Weigh the calculated mass of solute accurately.
2. Dissolve it in a small volume of distilled water in a beaker, stirring.
3. Transfer the solution into a volumetric flask, rinsing the beaker and stirring rod so no solute is left behind.
4. Add distilled water until the bottom of the meniscus sits on the graduation mark.
5. Stopper and invert several times to mix thoroughly.

Worked example 1 (easy), sodium hydroxide

Find the mass of sodium hydroxide, NaOH, needed to prepare 250 cm³ of 0.1 mol dm⁻³ solution. (M = 40 g mol⁻¹.)

• Step 1, volume: 250 cm³ = 0.25 dm³.
• Step 2, moles: n = 0.1 x 0.25 = 0.025 mol.
• Step 3, mass: m = 0.025 x 40 = 1.0 g.

Worked example 2 (medium), sodium carbonate

Find the mass of sodium carbonate, Na₂CO₃, needed for 500 cm³ of 0.2 mol dm⁻³ solution. (M = 106 g mol⁻¹.)

• Step 1, volume: 500 cm³ = 0.5 dm³.
• Step 2, moles: n = 0.2 x 0.5 = 0.1 mol.
• Step 3, mass: m = 0.1 x 106 = 10.6 g.

Worked example 3 (SPM level), a primary standard

Find the mass of anhydrous sodium carbonate needed to prepare 250 cm³ of a 0.05 mol dm⁻³ standard solution. (M = 106 g mol⁻¹.)

• Step 1, volume: 250 cm³ = 0.25 dm³.
• Step 2, moles: n = 0.05 x 0.25 = 0.0125 mol.
• Step 3, mass: m = 0.0125 x 106 = 1.325 g.

You weigh 1.325 g, dissolve, transfer with rinsings, and make up to 250 cm³ in the volumetric flask.

Common traps

• Making up to the mark in a beaker instead of a volumetric flask; only the flask gives an exact volume.
• Forgetting to convert cm cubed to dm cubed before finding the moles.
• Leaving out the rinsings, so some solute never reaches the flask.
• Using a hydrated formula mass when the question specifies the anhydrous solid, or vice versa.
• Reading the meniscus from the top rather than the bottom.

How we help

Our teachers rehearse the calculation and the full method together, so you can both compute 1.325 g and describe every step the marking scheme wants. Lessons are one-to-one, taught in English, from RM50 per hour, with a paid one-hour trial so you can try it first. A secure standard-solution routine feeds straight into titration, which is where these marks are most often won or lost in SPM Chemistry.