Calculations in analysis: titration and mass analysis
Answers are for parents and teachers.
1
2.03 g of MgCl₂·xH₂O loses 1.08 g of water (paper only; Mg = 24, Cl = 35.5). Find x and the percentage of water.
Dry salt 0.95 g; M(MgCl₂) = 95 g/mol; n = 0.0100 mol; water 1.08 ÷ 18 = 0.0600 mol; x = 6. Water = 1.08 ÷ 2.03 × 100 = 53.2 %.
2
25.0 mL of Ba(OH)₂ solution is neutralised by 20.0 mL of 0.100 M HCl (Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O). Find the concentration of Ba(OH)₂.
Acid na = 2, base nb = 1: 0.100 × 20.0 ÷ 2 = Mb × 25.0 ÷ 1, so Mb = 0.0400 mol/L.
3
2.00 g of impure KOH needs 30.0 mL of 0.500 M HCl (K = 39, O = 16, H = 1). Find the percentage of KOH.
n(HCl) = 0.0150 mol = n(KOH); m = 0.0150 × 56 = 0.84 g; 0.84 ÷ 2.00 × 100 = 42 %.
4
A sample of (NH₄)₂SO₄ (M = 132 g/mol) gives 1.398 g of BaSO₄ (M = 233 g/mol) with excess BaCl₂. Find the mass of (NH₄)₂SO₄ and the mass of sulphate ion SO₄²⁻ (M = 96 g/mol).
n(BaSO₄) = 1.398 ÷ 233 = 0.00600 mol; m((NH₄)₂SO₄) = 0.00600 × 132 = 0.792 g; m(SO₄²⁻) = 0.00600 × 96 = 0.576 g.
5
A student titrates 40.0 mL of H₂SO₄ with 30.0 mL of 0.200 M NaOH and writes Ma = 0.200 × 30.0 ÷ 40.0 = 0.150 mol/L. Find the slip and the correct concentration.
He forgot the coefficients (na = 1, nb = 2). Ma × 40.0 ÷ 1 = 0.200 × 30.0 ÷ 2, so Ma = 0.0750 mol/L.