Preparing solutions accurately is one of the most basic laboratory skills, and one of the most common sources of error. This article works through the calculations you will use most often, with examples, and highlights the traps that catch experienced chemists as well as students.
Molarity
Molarity (M) is moles of solute per litre of solution:
Molarity = moles of solute ÷ volume of solution in litres
Since moles = mass ÷ molecular weight, the mass required to prepare a solution is:
Mass (g) = Molarity (mol/L) × Volume (L) × Molecular weight (g/mol)
Example 1: 250 mL of 0.5 M sodium chloride
The molecular weight of NaCl is 58.44 g/mol. Mass = 0.5 × 0.250 × 58.44 = 7.305 g. Weigh 7.305 g, dissolve in less than 250 mL of water, transfer to a 250 mL volumetric flask and make up to the mark.
Example 2: the hydrate trap
You need 500 mL of 0.1 M copper(II) sulfate, but the bottle contains the pentahydrate, CuSO₄·5H₂O, molecular weight 249.69 g/mol, not the anhydrous salt at 159.61 g/mol. Mass = 0.1 × 0.500 × 249.69 = 12.48 g. Using the anhydrous molecular weight by mistake would give 7.98 g of pentahydrate and a solution only about 0.064 M. Always check the CAS number and formula on the label and COA.
Correcting for assay
If the reagent assay is 98.5%, divide the calculated mass by 0.985. For accurate standards, also account for water content if the assay is reported on the dried or anhydrous basis.
Molarity of concentrated reagents
Concentrated acids and ammonia are sold by weight percent. To find molarity, use density:
Molarity = (density in g/mL × 1000 × weight fraction) ÷ molecular weight
| Reagent | Typical % w/w | Density (g/mL) | MW | Approx. molarity |
|---|---|---|---|---|
| Hydrochloric acid | 37 | 1.19 | 36.46 | 12.1 M |
| Sulfuric acid | 98 | 1.84 | 98.08 | 18.4 M |
| Nitric acid | 70 | 1.41 | 63.01 | 15.7 M |
| Acetic acid (glacial) | 99.8 | 1.05 | 60.05 | 17.4 M |
| Ammonia solution | 25 | 0.91 | 17.03 | 13.4 M |
Exact values depend on the lot assay and density stated on the COA and label.
Example 3: 37% hydrochloric acid
Molarity = (1.19 × 1000 × 0.37) ÷ 36.46 = 440.3 ÷ 36.46 = 12.08 M.
Dilutions: C1V1 = C2V2
When you dilute a solution, the amount of solute stays the same, so:
C1 × V1 = C2 × V2
where C1 and V1 are the concentration and volume of the stock, and C2 and V2 those of the diluted solution. Units must be consistent.
Example 4: 500 mL of 1 M HCl from 12.08 M stock
V1 = (1 × 500) ÷ 12.08 = 41.4 mL. Add about 300 mL of water to a 500 mL volumetric flask, add 41.4 mL of concentrated acid slowly with swirling, cool to room temperature, then make up to the mark. Always add acid to water, never water to acid, particularly with sulfuric acid, which releases a great deal of heat.
Serial dilutions
For very dilute standards, dilute in steps rather than pipetting tiny volumes. To make a 1 µg/mL standard from a 1000 µg/mL stock, dilute 1 mL to 100 mL (10 µg/mL), then 10 mL of that to 100 mL (1 µg/mL). Each step uses volumes that can be measured accurately.
Normality
Normality (N) is equivalents of solute per litre. The number of equivalents per mole depends on the reaction:
Normality = Molarity × n
- In acid-base reactions, n is the number of H⁺ ions donated or accepted per molecule. HCl: n = 1. H₂SO₄: n = 2. So 0.1 M H₂SO₄ is 0.2 N.
- In redox reactions, n is the number of electrons transferred per molecule. Permanganate in acid solution is reduced from Mn(VII) to Mn(II), so n = 5 and 0.02 M KMnO₄ is 0.1 N.
Because normality depends on the reaction, the same solution can have different normalities in different contexts. This is why IUPAC discourages normality in favour of molarity with the reaction specified, although it remains common in titrimetry and pharmacopoeial methods. Concentrated sulfuric acid at 18.4 M is 36.8 N as an acid.
Titration using normality
At the equivalence point, equivalents of titrant equal equivalents of analyte: N1 × V1 = N2 × V2. If 25.0 mL of an NaOH solution requires 21.6 mL of 0.1 N HCl, the NaOH normality is (0.1 × 21.6) ÷ 25.0 = 0.0864 N.
Percent concentrations
- % w/v: grams of solute per 100 mL of solution. A 5% w/v glucose solution contains 5 g per 100 mL.
- % w/w: grams of solute per 100 g of solution, as used for concentrated acids.
- % v/v: millilitres of solute per 100 mL of solution, as used for ethanol-water mixtures. Volumes are not strictly additive, so make up to final volume.
ppm and ppb
For dilute aqueous solutions, 1 ppm is approximately 1 mg/L and 1 ppb approximately 1 µg/L. To convert a molar concentration to mg/L, multiply by the molecular weight and by 1000. For example, 0.001 M NaCl is 0.001 × 58.44 × 1000 = 58.4 mg/L.
Checklist for accurate solutions
- Confirm the exact form, anhydrous or hydrate, salt or free base, from the CAS number.
- Correct for assay and water content where accuracy matters.
- Use calibrated balances and Class A volumetric glassware.
- Make up to volume at the temperature the flask is calibrated for.
- Standardise titrants against a primary standard rather than relying on the calculated value.
- Label solutions with concentration, date, preparer and expiry.
FAQs
How do I convert molarity to normality?
Multiply molarity by the number of equivalents per mole for the reaction: protons for acid-base reactions or electrons for redox reactions.
How much 37% HCl do I need for 1 L of 1 M acid?
Concentrated 37% HCl is about 12.1 M, so you need about 83 mL made up to 1 L, adding the acid to water.
Why does my solution concentration come out wrong?
Common causes are using the wrong molecular weight for a hydrate, ignoring assay or water content, and making up to volume at the wrong temperature.