Zadanie 1d

(i) [CH₃COOH] «$= 0.100\ \mathrm{mol\,dm^{-3}} \times \dfrac{18.4\ \mathrm{cm^3}}{25.0\ \mathrm{cm^3}}$» = 0.0736 «mol dm⁻³» ✔

mass of CH₃COOH «= 2.00 dm³ × 0.0736 mol dm⁻³ × 60.06 g mol⁻¹» = 8.84 «g» ✔

(ii) $K_\mathrm{a} = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}$ with [H⁺] = [CH₃COO⁻], so $[\ce{H+}] = \sqrt{K_\mathrm{a}\,[\ce{CH3COOH}]}$ ✔

[H⁺] «$= \sqrt{1.74 \times 10^{-5} \times 0.0736} = \sqrt{1.28 \times 10^{-6}}$» = 1.13 × 10⁻³ «mol dm⁻³» ✔

pH = 2.95 ✔

(iii) CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq) ✔

ethanoate is a proton acceptor «Brønsted–Lowry base», so the solution is alkaline AND pH > 7 ✔

(iv) the double / π bond «in the ethanoate ion» is delocalized

OR

resonance occurs ✔

each carbon–oxygen bond is shorter than the single C–O bond AND longer than the C=O bond «in ethanoic acid» / intermediate in length between a single and a double bond ✔

(i) M1: concentration of the ethanoic acid solution; M2: mass. Award [2] for a correct final answer; accept 8.8 g. Award [0] for 0.184 g (mass in 25.0 cm³ of solution). (ii) M1: expression for [H⁺] from K_a; M2: value of [H⁺]; M3: pH. Award [3] for a correct final answer; accept 2.94–2.96. ECF from (i). (iii) M1: hydrolysis equation with equilibrium sign (accept arrows); the neutralization equation of the titration earns nothing. M2: base AND pH > 7; M2 may be implied by the equation. M2 is also earned by identifying the salt as one formed from a strong base and a weak acid, together with pH > 7. (iv) M1: delocalization or resonance; M2: comparison of bond lengths. Accept a drawing of the resonance structures for M1.