Question 21M.2.hl.TZ2.5

CH₃CH₂OH (l) + 3O₂(g) → 2CO₂(g) + 3H₂O (g) ✓

«n = $\frac{56.00 \mathrm{g}}{46.08 \mathrm{g} {\mathrm{mol}}^{-1}}$ =» 1.215 «mol» ✓

«1.215mol × (−1367 kJ mol⁻¹) =» −1661 «kJ» ✓

Award [2] for correct final answer.

Award [1 max] for “«+»1661 «kJ»”.

ethanal AND distillation ✓

ethanoic acid AND reflux «followed by distillation» ✓

Award [1] for both products OR both methods.

Equation:
CH₃CH₂OH + HCOOH $\rightleftharpoons$ HCOOCH₂CH₃ + H₂O ✓

Product name:
ethyl methanoate ✓

Accept equation without equilibrium arrows.

Accept equation with molecular formulas (C₂H₆O + CH₂O₂ $\rightleftharpoons$ C₃H₆O₂ + H₂O) only if product name is correct.

increasing S-shape pH curve ✓

pK_a: pH at half neutralization/equivalence ✓

M1: Titration curve must show buffer region at pH <7 and equivalence at pH >7.

Ignore other parts of the curve, i.e., before buffer region, etc.

Accept curve starting from where two axes meet as pH scale is not specified.

phenolphthalein
OR
phenol red ✓

Alternative 1:
K_a = $\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HCOO}}^{-}\right]}{\left[\mathrm{HCOOH}\right]}$
OR
[HCOOH] = $\frac{{\left({10}^{-4.12}\right)}^2}{{10}^{-3.75}}$✓

«[HCOOH] =» 3.24 × 10⁻⁵ «mol dm⁻³» ✓

Alternative 2:
«pH = pK_a + log $\frac{\left[{\mathrm{HCOO}}^{-}\right]}{\left[\mathrm{HCOOH}\right]}$»
4.12 = 3.75 + log$\frac{{10}^{-4.12}}{\left[\mathrm{HCOOH}\right]}$ ✓

«[HCOOH] =» 3.24 × 10⁻⁵ «mol dm⁻³» ✓

Award [2] for correct final answer.

Sodium methanoate: basic

Ammonium chloride: acidic

Sodium nitrate: neutral ✓ ✓

Award [2] for three correct.

Award [1] for two correct.