The problems are on the net ionic equation worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A net ionic equation shows only the species that change, leaving out spectator ions. These worked answers correct the 20 problems of the AP Chemistry worksheet on net ionic equation, from soluble or insoluble to identify four solutions: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the method, the substitution and the result with units. For equations, check that both atoms and charges balance. Allow small differences in the last digit because of rounding of molar masses.
Level: AP Chemistry · Curriculum: AP Chem Unit 4
Problem 1 Soluble or insoluble
NaCl soluble; AgCl insoluble; BaSO₄ insoluble; K₂CO₃ soluble (group 1); Mg(OH)₂ insoluble; NH₄NO₃ soluble.Problem 2 Find the spectators
1. Ag⁺(aq) + Cl⁻(aq) → AgCl(s).2. Na⁺ and NO₃⁻.
Problem 3 Oxidation numbers
1. x + 4(−2) = −2, so S = +6.2. 2x + 7(−2) = −2, so Cr = +6.
3. x + 4(+1) = +1, so N = −3.
4. x + 3(−2) = −1, so Cl = +5.
Problem 4 Oxidised or reduced
Mg goes from 0 to +2: Mg is oxidised and is the reducing agent. Cu²⁺ goes from +2 to 0: Cu²⁺ is reduced (oxidising agent).Problem 5 Reading a titration curve
1. About pH 1 (0.100 M strong acid).2. About 25 mL, in the middle of the steep rise.
3. pH 7; phenolphthalein or bromothymol blue, both of which change colour within the steep part.
Problem 6 Volume added from a burette
1. 21.60 − 0.30 = 21.30 mL.2. n(NaOH) = 0.02130 × 0.100 = 2.130 × 10⁻³ mol = n(HCl); c = 2.130 × 10⁻³ / 0.01000 = 0.213 mol/L.
Problem 7 Three kinds of equation
1. Na₂CO₃(aq) + CaCl₂(aq) → CaCO₃(s) + 2NaCl(aq)2. 2Na⁺ + CO₃²⁻ + Ca²⁺ + 2Cl⁻ → CaCO₃(s) + 2Na⁺ + 2Cl⁻
3. Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)
Problem 8 Neutralisation equations
1. H⁺(aq) + OH⁻(aq) → H₂O(l).2. CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l); the weak acid is not split into ions, and Na⁺ is a spectator.
Problem 9 Predict the precipitate
1. Yes, BaSO₄.2. No, all products are soluble.
3. Yes, Fe(OH)₃.
4. Yes, ZnS.
Problem 10 Mass of precipitate
n(Ag⁺) = 0.0250 × 0.200 = 5.00 × 10⁻³ mol = n(AgCl)m = 5.00 × 10⁻³ × 143.32 = 0.717 g.
Problem 11 Limiting reagent in solution
n(Pb²⁺) = 6.00 × 10⁻³ mol; n(I⁻) = 15.0 × 10⁻³ mol. Pb²⁺ + 2I⁻ needs 12.0 × 10⁻³ mol of I⁻.I⁻ is in excess, so Pb(NO₃)₂ is limiting.
m = 6.00 × 10⁻³ × 461.0 = 2.77 g.
Problem 12 Percent yield
Theoretical = 2.766 g (unrounded). Percent yield = 2.45 / 2.766 × 100 = 88.6 %.Problem 13 Sulfuric acid titration
n(NaOH) = 0.03050 × 0.1000 = 3.050 × 10⁻³ moln(H₂SO₄) = 3.050 × 10⁻³ / 2 = 1.525 × 10⁻³ mol
c = 1.525 × 10⁻³ / 0.02500 = 0.0610 mol/L.
Problem 14 Half-equations
1. Oxidation: Zn → Zn²⁺ + 2e⁻. Reduction: Ag⁺ + e⁻ → Ag.2. Zn gives 2 electrons and each Ag⁺ accepts only 1, so electrons balance with 2 Ag⁺.
Problem 15 Dissolving ionic compounds
1. Na₃PO₄(s) → 3Na⁺(aq) + PO₄³⁻(aq).2. Al(NO₃)₃(s) → Al³⁺(aq) + 3NO₃⁻(aq).
In 0.10 M Na₃PO₄: [Na⁺] = 0.30 M, [PO₄³⁻] = 0.10 M.
Problem 16 Concentration of vinegar
1. n(NaOH) = 0.01820 × 0.1000 = 1.820 × 10⁻³ mol = n(acid); c = 1.820 × 10⁻³ / 0.02500 = 0.0728 mol/L.2. Dilution factor 10: 0.728 mol/L.
3. 0.728 × 60.05 = 43.7 g/L, so 4.37 g per 100 mL.
Problem 17 Redox titration of iron
1. n(MnO₄⁻) = 0.02040 × 0.02000 = 4.080 × 10⁻⁴ mol; n(Fe²⁺) = 5 × 4.080 × 10⁻⁴ = 2.040 × 10⁻³ mol; c = 2.040 × 10⁻³ / 0.02500 = 0.0816 mol/L.2. Mn: +7 to +2 (reduced); Fe: +2 to +3 (oxidised).
Problem 18 Ions left after precipitation
1. Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).2. n = 5.00 × 10⁻³ mol of each, so m = 5.00 × 10⁻³ × 233.39 = 1.17 g.
3. Total volume 0.1000 L. n(Na⁺) = 2 × 5.00 × 10⁻³ = 1.00 × 10⁻² mol; n(Cl⁻) the same; each concentration is 0.100 mol/L.
Problem 19 Analyse titration data
1. Use concordant results (within 0.10 mL): trials 2, 3 and 4. The rough trial is discarded because the end point was overshot.2. Mean = (21.60 + 21.55 + 21.65)/3 = 21.60 mL; n(NaOH) = 0.02160 × 0.1050 = 2.268 × 10⁻³ mol; c(HCl) = 2.268 × 10⁻³ / 0.02500 = 0.0907 mol/L.
Problem 20 Identify four solutions
1 = Pb(NO₃)₂ (the only one that precipitates with two others); 2 = KI (yellow PbI₂); 3 = Na₂SO₄ (white PbSO₄); 4 = NaNO₃ (no reaction).Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s)
