🌳 Concept tree

• Lesson 3 Quantitative Chemical Analysis – Volumetric and Gravimetric Methods
• ├── Quantitative analysis — volumetric analysis
• ├── End point and indicators
• ├── Procedure (NaOH titrated with HCl)
• ├── Law of titration
• ├── Quantitative analysis — gravimetric analysis
• ├── A. Volatilization method
• ├── B. Precipitation method
• ├── Choosing the method
• └── Summary

• For the reaction n_a acid + n_b base → products:

Ma × Va / na = Mb × Vb / nb

• (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 = Ma Va/1000; since 1 mol acid reacts with 1 mol base, moles of base in the sample equal that number; Mb = moles × 1000 / Vb.
• 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 %.

• 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.