Solution Stoichiometry and Molar Volumes

Solution stoichiometry applies mole ratios to reactions in solution, using molarity (moles of solute per liter of solution) to convert between the volume of a solution and the number of moles: moles = molarity × volume in liters. In brief: 25.00 mL of 0.100 mol/L HCl contains 0.02500 L × 0.100 mol/L = 2.50 × 10⁻³ mol of HCl, which reacts with exactly 2.50 × 10⁻³ mol of NaOH, so it needs 12.50 mL of 0.200 mol/L NaOH.

Most reactions in laboratories, industry and living organisms take place in solution. Solutions are measured by volume, so molarity takes the place of molar mass as the conversion factor between a measured quantity and moles. This article explains molarity, how to set up solution stoichiometry problems, how titrations work, how to treat the volume of gases formed from solutions with the molar volume, shows tables, works through examples with precipitation, titration and gas evolution and gives the history of volumetric analysis.

What is molarity?

Molarity (M) is the number of moles of solute per liter of solution: M = n ÷ V, with units mol/L. A 2.0 mol/L solution (2.0 M) contains 2.0 mol of solute in each liter of solution. Rearranged: n = M × V, where V must be in liters. 1 L = 1,000 mL, so 25.0 mL = 0.0250 L. The volume of a solution, instead of the mass, serves as the measurable quantity, and the molarity acts like a “molar volume” of the solution: it states how many moles are in each liter.

How do you solve a solution stoichiometry problem?

1. Write the balanced equation (a net ionic equation is often the clearest).
2. Convert the given quantity to moles: n = M × V for a solution, n = m ÷ M for a solid and n = V ÷ 22.4 L/mol for a gas at STP.
3. Use the mole ratio to find the moles of the wanted substance.
4. Convert moles to the required quantity: volume of a solution (V = n ÷ M), mass (m = n × M) or volume of a gas.

How do you find the concentration of ions?

A soluble ionic compound dissociates completely, and the concentration of each ion is the concentration of the compound times the number of ions in the formula. For 0.100 mol/L CaCl₂: [Ca²⁺] = 0.100 mol/L and [Cl⁻] = 2 × 0.100 = 0.200 mol/L. For 0.050 mol/L Al₂(SO₄)₃: [Al³⁺] = 0.100 mol/L and [SO₄²⁻] = 0.150 mol/L.

Conversion factors in stoichiometry

Worked examples

1. Volume of base for neutralization. HCl + NaOH → NaCl + H₂O. What volume of 0.200 mol/L NaOH is needed for 25.00 mL of 0.100 mol/L HCl? n(HCl) = 0.02500 × 0.100 = 2.50 × 10⁻³ mol. n(NaOH) = 2.50 × 10⁻³ mol. V = 2.50 × 10⁻³ ÷ 0.200 = 0.01250 L = 12.50 mL.

2. Precipitate. Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃. What mass of PbI₂ forms when 25.0 mL of 0.200 mol/L Pb(NO₃)₂ reacts with excess KI? n(Pb²⁺) = 0.0250 × 0.200 = 5.00 × 10⁻³ mol. n(PbI₂) = 5.00 × 10⁻³ mol. m = 5.00 × 10⁻³ × 461.0 = 2.31 g.
3. Gas from solution. Zn + 2HCl → ZnCl₂ + H₂. What volume of H₂ at STP forms when 50.0 mL of 2.00 mol/L HCl reacts with excess zinc? n(HCl) = 0.0500 × 2.00 = 0.100 mol. n(H₂) = 0.100 ÷ 2 = 0.0500 mol. V = 0.0500 × 22.4 = 1.12 L.
4. Mass of solid needed. What mass of Na₂CO₃ reacts with 40.0 mL of 0.500 mol/L HCl, Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂? n(HCl) = 0.0400 × 0.500 = 0.0200 mol. n(Na₂CO₃) = 0.0100 mol. m = 0.0100 × 105.99 = 1.06 g.
5. Diprotic acid. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. How much 0.250 mol/L NaOH neutralizes 20.0 mL of 0.150 mol/L H₂SO₄? n(H₂SO₄) = 0.0200 × 0.150 = 3.00 × 10⁻³ mol. n(NaOH) = 6.00 × 10⁻³ mol. V = 6.00 × 10⁻³ ÷ 0.250 = 0.0240 L = 24.0 mL.
6. Concentration after mixing. 50.0 mL of 0.100 mol/L AgNO₃ is added to 50.0 mL of 0.200 mol/L NaCl. Ag⁺: 5.00 × 10⁻³ mol. Cl⁻: 0.0100 mol. Ag⁺ + Cl⁻ → AgCl. Chloride in excess: 0.0100 − 0.00500 = 5.00 × 10⁻³ mol in 100.0 mL (0.1000 L): [Cl⁻] = 0.0500 mol/L. The mass of AgCl is 5.00 × 10⁻³ × 143.32 = 0.717 g.

How does a titration work?

A titration measures the volume of a solution of known concentration (the titrant, delivered from a burette) needed to react completely with a measured volume of another solution. The point at which the reactants are in exact stoichiometric proportion is the equivalence point, which is detected by a color change of an indicator (the end point) or by a pH meter.

Example. A 25.00 mL sample of sulfuric acid is titrated with 0.1000 mol/L NaOH, and the end point is reached after 31.45 mL. n(NaOH) = 0.03145 × 0.1000 = 3.145 × 10⁻³ mol. From H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O: n(H₂SO₄) = 1.5725 × 10⁻³ mol. The concentration is 1.5725 × 10⁻³ ÷ 0.02500 = 0.06290 mol/L.

Another example. 20.00 mL of HCl needs 18.62 mL of 0.1000 mol/L NaOH. n(NaOH) = 1.862 × 10⁻³ mol = n(HCl). [HCl] = 1.862 × 10⁻³ ÷ 0.02000 = 0.09310 mol/L.

What happens when a solution is diluted?

Adding solvent does not change the number of moles of solute: n = M₁V₁ = M₂V₂. To make 250.0 mL of 0.100 mol/L solution from a 2.00 mol/L stock, V₁ = 0.100 × 250.0 ÷ 2.00 = 12.5 mL of the stock solution diluted to 250.0 mL.

How did volumetric analysis develop?

Volumetric analysis, in which a measured volume of a standard solution is used to determine the amount of a substance, started in the late 18th century. François Descroizilles developed the burette around 1791 for bleach analysis. Joseph Louis Gay-Lussac improved the method in the 1820s, and Karl Friedrich Mohr’s textbook of 1855 made titrations a standard laboratory method. The concept of molarity became general in the 20th century.

Which mistakes are common?

Mistake 1: Using milliliters in n = MV

Convert to liters first, or use millimoles (M × mL).

Mistake 2: Dividing by the volume of the solvent

Molarity is per liter of solution, not per liter of solvent.

Mistake 3: Ignoring the mole ratio in a titration

A diprotic acid needs 2 mol of NaOH per mole of acid.

Mistake 4: Adding the concentrations of mixed solutions

Calculate the total moles and divide by the total volume after mixing.