Molarity, Dilution and Solution Stoichiometry

Molarity counts moles per liter of solution

Molarity, also called amount concentration, is the amount of solute in moles divided by the volume of solution in liters: c = n/V, in mol/L, often written M. A 0.250 mol/L glucose solution contains 0.250 mol of glucose in every liter of the final solution. Square brackets denote the molar concentration of a species, as in [Cl⁻].

The volume is the volume of solution, not of solvent. Dissolving a solute in exactly 1.000 L of water generally gives a total volume that is not 1.000 L, so solutions are made up to a calibration mark in a volumetric flask rather than by adding a measured volume of water.

For a strong electrolyte, ion concentrations follow from the formula. A 0.150 mol/L CaCl₂ solution is 0.150 mol/L in Ca²⁺ and 0.300 mol/L in Cl⁻, because each formula unit releases two chloride ions. Molarity also changes slightly with temperature, since solutions expand when warmed.

Preparing a solution from a weighed solid

The amount needed is n = c × V, and the mass to weigh is n × M, where M is the molar mass. Weigh the solid, dissolve it in less water than the final volume, transfer it to a volumetric flask while rinsing the beaker into the flask, dilute to the mark, and mix by inverting.

Worked example: prepare 250.0 mL of 0.0800 mol/L copper(II) sulfate from blue crystals of copper(II) sulfate pentahydrate, CuSO₄·5H₂O. With IUPAC standard atomic weights rounded to two decimals (Cu 63.55, S 32.06, O 16.00, H 1.01), the molar mass of the hydrate is 63.55 + 32.06 + 4(16.00) + 5(18.02) = 249.71 g/mol.

n = 0.0800 mol/L × 0.2500 L = 0.02000 mol. Mass = 0.02000 mol × 249.71 g/mol = 4.994 g, so weigh 4.99 g of the pentahydrate. Each formula unit still supplies one Cu²⁺ and one SO₄²⁻, but the water of hydration adds to the mass you weigh. If you used the anhydrous molar mass, 159.61 g/mol, and weighed 3.19 g of the hydrate, the solution would be only 0.0511 mol/L, about 36% too dilute.

Diluting a stock solution with C₁V₁ = C₂V₂

Adding solvent changes the volume but not the amount of solute. Since n = cV before and after dilution, C₁V₁ = C₂V₂, where 1 refers to the stock and 2 to the diluted solution. Any volume unit works if both volumes use the same one, because the units cancel.

Worked example: you need 500.0 mL of 0.360 mol/L sodium chloride and have a 2.40 mol/L stock. V₁ = C₂V₂ ÷ C₁ = (0.360 mol/L × 500.0 mL) ÷ 2.40 mol/L = 75.0 mL. Check: the stock is 2.40 ÷ 0.360 = 6.67 times as concentrated, so you need about one-seventh of the final volume, and 500.0 mL ÷ 6.67 = 75.0 mL.

Measure 75.0 mL of stock into a 500-mL volumetric flask and dilute to the mark. Do not simply add 425.0 mL of water: volumes of solutions are not strictly additive, and C₂ refers to the final volume. When diluting a concentrated acid, add the acid to water rather than water to the acid, because the heat released can cause spattering.

Solution stoichiometry: from volume to moles and back

In reactions between solutions, molarity converts volume to moles through n = c × V. The mole ratio from the balanced equation then applies exactly as it does in mass-based stoichiometry, and a second molarity converts back to a volume if needed.

Worked example: what volume of 0.125 mol/L barium chloride precipitates all the sulfate from 40.0 mL of 0.0850 mol/L sodium sulfate, and what mass of barium sulfate forms? The net ionic equation is Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). n(SO₄²⁻) = 0.0850 mol/L × 0.0400 L = 3.40 × 10⁻³ mol. The ratio is 1:1, and each BaCl₂ supplies one Ba²⁺, so V = 3.40 × 10⁻³ mol ÷ 0.125 mol/L = 0.0272 L, or 27.2 mL.

The molar mass of BaSO₄ (Ba 137.33, S 32.06, O 16.00) is 233.39 g/mol, so the mass formed is 3.40 × 10⁻³ mol × 233.39 g/mol = 0.794 g. Each answer has three significant figures, set by 0.0850 mol/L, 40.0 mL and 0.125 mol/L.

Titration: turning a buret reading into a concentration

In a titration, a solution of known concentration, the titrant, is added from a buret to a measured volume of sample until the reaction is complete. The equivalence point is where the amount of titrant added matches the sample in the stoichiometric ratio. The endpoint is where the indicator changes color, and a suitable indicator makes the two nearly coincide.

Worked example: a 20.00 mL sample of barium hydroxide solution of unknown concentration requires 28.36 mL of 0.1208 mol/L hydrochloric acid to reach a phenolphthalein endpoint. Ba(OH)₂(aq) + 2 HCl(aq) → BaCl₂(aq) + 2 H₂O(l). n(HCl) = 0.1208 mol/L × 0.02836 L = 3.426 × 10⁻³ mol. n(Ba(OH)₂) = 3.426 × 10⁻³ mol ÷ 2 = 1.713 × 10⁻³ mol. c(Ba(OH)₂) = 1.713 × 10⁻³ mol ÷ 0.02000 L = 0.08565 mol/L, to four significant figures.

The factor of 2 matters. Treating the reaction as 1:1 would give 0.1713 mol/L, twice the true value, which is why C₁V₁ = C₂V₂ should not be used for titrations. The titrant's own concentration is established beforehand by standardization, titrating it against a weighed primary standard; sodium carbonate is a common choice for acids.

Concentration mistakes and how to spot them

Most concentration errors are unit or setup errors. Writing units on every quantity, and checking that the result moves in the expected direction, catches nearly all of them.

A diluted solution must be less concentrated than its stock, a mass weighed for a hydrate must be larger than for the anhydrous salt, and a titrant volume should fall within the buret's range. A result that breaks any of these deserves a second look.

• Using milliliters in n = c × V with c in mol/L. Convert to liters first; mL × mol/L does not give mol.
• Using the volume of solvent instead of the final volume of solution.
• Using the anhydrous molar mass for a hydrate.
• Applying C₁V₁ = C₂V₂ to a reaction. It describes dilution only; for reactions, go through moles and the mole ratio.
• Forgetting ion multiplicity, such as treating 0.150 mol/L CaCl₂ as 0.150 mol/L in Cl⁻.
• Adding the final volume of water instead of diluting to the final volume.