Lesson 3: Quantitative Chemical Analysis - Volumetric and Gravimetric Methods

• Volumetric analysis is based on measuring the volumes of the substances being analysed: the volume of a solution of known concentration (standard solution) which is completely consumed by a known volume of the analyte gives the concentration of the analyte.
• The process is titration.
• To choose the standard solution we must know the suitable reaction between the two solutions.
• It may be:

1. Neutralization (acid–base) — for acids and bases;
2. Oxidation–reduction — for oxidizing or reducing substances;
3. Precipitation — for substances that give sparingly soluble products.

If the analyte is an acid, a standard alkali (NaOH, or Na₂CO₃) is used; if it is basic, a standard acid is used.

• The end point is the point at which the neutralization reaction is just complete.
• An indicator (a substance that changes colour with the medium) shows it.

For the titration of a strong acid with a strong base (HCl with NaOH) any of these indicators can be used, because the sharp change in pH at the end point is much wider than the transition range of the indicator.

• Transfer a measured volume (e.g. 25 mL) of the alkali to a conical flask with a pipette; add two drops of indicator (methyl orange or phenolphthalein); fill the burette with the standard 0.1 M acid; add the acid gradually, swirling the flask, in small quantities until the alkali has been completely neutralized as shown by the change in the colour of the indicator.
• Read the volume used.
• A good result needs: a clean burette rinsed with the acid, a pipette rinsed with the alkali, adding the acid drop by drop near the end point, and repeating the titration until two readings agree.

For the reaction n_a acid + n_b base → products:

M_a × V_a / n_a = M_b × V_b / n_b

(M = molarity, V = volume (same unit on both sides), n = number of moles of acid/base in the balanced equation).

Reasoning for HCl + NaOH: moles of acid used = M_a V_a/1000; since 1 mol acid reacts with 1 mol base, moles of base in the sample equal that number; M_b = moles × 1000 / V_b.

Worked example 1. 20 mL of Ca(OH)₂ solution needs 30 mL of 0.04 M HCl for complete neutralization. 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O, so n_a = 2, n_b = 1. 0.04 × 30 / 2 = M_b × 20 / 1 → M_b = 0.6/20 = 0.03 M.

Worked example 2 (percentage purity). A 0.5 g mixture of NaOH and NaCl needs 25 mL of 0.2 M HCl. n(HCl) = 0.2 × 25/1000 = 0.005 mol = n(NaOH). Mass of NaOH = 0.005 × 40 = 0.2 g. % NaOH = 0.2/0.5 × 100 = 40 %. (NaCl does not react with HCl.)

Worked example 3 (sulphuric acid, a dibasic acid). 25 mL of sodium hydroxide solution needs 17.5 mL of 0.08 M H₂SO₄ for neutralization. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so n_a = 1, n_b = 2. 0.08 × 17.5 / 1 = M_b × 25 / 2 → M_b = 0.08 × 17.5 × 2 / 25 = 0.112 M. Concentration in g/L = 0.112 × 40 = 4.48 g/L.

Worked example 4 (volume of acid required). What volume of 0.25 M HCl is needed to react completely with 30 mL of 0.15 M Na₂CO₃ solution (the products being NaCl, water and CO₂)? 2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂, so n_a = 2, n_b = 1. 0.25 × V_a / 2 = 0.15 × 30 / 1 → V_a = 0.15 × 30 × 2 / 0.25 = 36 mL.

Worked example 5 (mass of base). 40 mL of potassium hydroxide solution needs 22 mL of 0.1 M HCl. n(HCl) = 0.1 × 22/1000 = 2.2 × 10⁻³ mol = n(KOH). Mass of KOH = 2.2 × 10⁻³ × 56 = 0.1232 g in the 40 mL.

Worked example 6 (purity of sodium carbonate). 1.2 g of impure Na₂CO₃ is dissolved in water and needs 40 mL of 0.5 M HCl to reach the end point (methyl orange). n(HCl) = 0.5 × 40/1000 = 0.02 mol. By 2HCl + Na₂CO₃ → …, n(Na₂CO₃) = 0.01 mol. Mass = 0.01 × 106 = 1.06 g. Purity = 1.06/1.2 × 100 = 88.3 %.

• This method depends on separating the analyte and finding its mass, then using a chemical calculation (stoichiometry) to find the quantity of analyte.
• Two approaches:

The element or compound to be determined is converted to a volatile form; its mass is found either by collecting the volatile substance and weighing it or by the loss in mass of the original sample.

Worked example (water of crystallization). 3.05 g of hydrated barium chloride BaCl₂·xH₂O is heated strongly to constant mass 2.60 g (Ba = 137, Cl = 35.5, H = 1, O = 16).

• Mass of water = 3.05 − 2.60 = 0.45 g; % water = 0.45/3.05 × 100 ≈ 14.75 %.
• Moles BaCl₂ = 2.60/208 = 0.0125 mol; moles H₂O = 0.45/18 = 0.025 mol.
• x = 0.025/0.0125 = 2 → BaCl₂·2H₂O.

Worked example (copper sulphate). 2.495 g of hydrated copper sulphate CuSO₄·xH₂O lose water on heating and leave 1.595 g of anhydrous salt (Cu = 63.5, S = 32, O = 16, H = 1). Mass of water = 0.900 g. Moles CuSO₄ = 1.595/159.5 = 0.0100 mol; moles H₂O = 0.900/18 = 0.0500 mol; x = 5 → CuSO₄·5H₂O. The percentage of water of crystallization = 0.9/2.495 × 100 = 36.1 %.

Worked example (loss of a gas). 5.0 g of an impure limestone sample lose 1.76 g when strongly heated: CaCO₃ → CaO + CO₂. Moles of CO₂ = 1.76/44 = 0.04 mol = moles of CaCO₃ → mass 0.04 × 100 = 4.0 g; purity = 4.0/5.0 × 100 = 80 %.

Percentage of water in a hydrate. For Na₂SO₄·10H₂O (M = 142 + 180 = 322 g/mol): % water = 180/322 × 100 = 55.9 %.

• The analyte is precipitated in the form of a pure, sparingly soluble compound of constant, known chemical composition.
• The precipitate is separated by filtration through ashless filter paper (a paper which leaves no ash when burnt), the paper with the precipitate is transferred to an ignition crucible and ignited until the paper is completely burnt off (its constituents volatilize), and the mass of the precipitate that remains is weighed and used to find the mass of the analyte.
• Example: barium precipitated as BaSO₄.

Worked example. Excess sodium sulphate is added to a solution containing BaCl₂; the BaSO₄ precipitate weighs 1.165 g (Ba = 137, S = 32, O = 16, Cl = 35.5). BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl 208 g BaCl₂ → 233 g BaSO₄ x → 1.165 g ⇒ x = 208 × 1.165/233 = 1.04 g BaCl₂.

Worked example (percentage of chloride). 1.5 g of impure NaCl gives 2.87 g AgCl (Ag = 108, Cl = 35.5): moles AgCl = 2.87/143.5 = 0.02 mol, so 0.02 mol Cl⁻ = 0.71 g Cl; percentage of chlorine = 0.71/1.5 × 100 = 47.3 %. (Mass of NaCl = 0.02 × 58.5 = 1.17 g, i.e. 78 % purity.)

Worked example (sulphate in a sample). A 0.60 g sample containing sodium sulphate is dissolved and excess BaCl₂ added; the dried precipitate of BaSO₄ weighs 0.699 g. Moles BaSO₄ = 0.699/233 = 0.00300 mol. Sulphate present = 0.00300 × 96 = 0.288 g. Mass of Na₂SO₄ = 0.00300 × 142 = 0.426 g; purity = 0.426/0.60 × 100 = 71.0 %.

Worked example (iron as the oxide). The iron in a sample is precipitated as Fe(OH)₃ and ignited to Fe₂O₃, which weighs 0.800 g (Fe = 56, O = 16). Fe₂O₃ (160 g) contains 112 g Fe, so the mass of iron = 0.800 × 112/160 = 0.560 g.

• If the substance can be turned into a gas by heating, or contains water of crystallization, use volatilization.
• If the analyte forms a pure, insoluble compound of known formula, use precipitation.
• If the analyte is an acid or a base (or an oxidising or reducing agent) in solution, use titration.