DP Chemistry (last assessment 2024)

Question 18N.2.hl.TZ0.6

Date	November 2018	Marks available	[Maximum mark: 15]	Reference code	18N.2.hl.TZ0.6
Level	hl	Paper	2	Time zone	TZ0
Command term	Calculate, Deduce, Determine, Draw, Explain, Sketch, State	Question number	6	Adapted from	N/A

[Maximum mark: 15]

18N.2.hl.TZ0.6

Butanoic acid, CH₃CH₂CH₂COOH, is a weak acid and ethylamine, CH₃CH₂NH₂, is a weak base.

(a.i)

State the equation for the reaction of each substance with water.

[2]

Markscheme

Butanoic acid:
CH₃CH₂CH₂COOH (aq) + H₂O (l) $\rightleftharpoons$ CH₃CH₂CH₂COO⁻ (aq) + H₃O⁺ (aq) ✔

Ethylamine:
CH₃CH₂NH₂ (aq) + H₂O (l) $\rightleftharpoons$ CH₃CH₂NH₃⁺(aq) + OH⁻ (aq) ✔

(a.ii)

Draw a diagram showing the delocalization of electrons in the conjugate base of butanoic acid.

[1]

Markscheme

Diagram showing:
dotted line along O–C–O AND negative charge

Accept correct diagrams with pi clouds.

(a.iii)

Deduce the average oxidation state of carbon in butanoic acid.

[1]

Markscheme

–1 ✔

(b.i)

A 0.250 mol dm⁻³ aqueous solution of butanoic acid has a concentration of hydrogen ions, [H⁺], of 0.00192 mol dm⁻³. Calculate the concentration of hydroxide ions, [OH⁻], in the solution at 298 K.

[1]

Markscheme

« $\frac{{1.00 \times {{10}^{ - 14}}\,{\text{mo}}{{\text{l}}^2}\,{\text{d}}{{\text{m}}^{ - 6}}}}{{0.00192\,{\text{mol}}\,{\text{d}}{{\text{m}}^{ - 3}}}}$ » = 5.21 × 10^–12 «mol dm^–3» ✔

(b.ii)

Determine the pH of a 0.250 mol dm⁻³ aqueous solution of ethylamine at 298 K, using section 21 of the data booklet.

[3]

Markscheme

«pK_b = 3.35, K_b = 10^–3.35 = 4.5 × 10^–4»

«C₂H₅NH₂ + H₂O $\rightleftharpoons$ C₂H₅NH₃⁺ + OH^–»

K_b = $\frac{{\left[ {{\text{O}}{{\text{H}}^--}} \right]\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{H}}_{\text{3}}}^{\text{ + }}} \right]}}{{\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{H}}_{\text{2}}}} \right]}}$

«K_b =» 4.5 × 10^–4 = $\frac{{\left[ {{\text{O}}{{\text{H}}^-}} \right]\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{H}}_{\text{3}}}^{\text{ + }}} \right]}}{{0.250}}$

«K_b =» 4.5 × 10^–4 = $\frac{{{x^2}}}{{0.250}}$ ✔

« x = [OH^–] =» 0.011 «mol dm^–3» ✔

«pH = –log $\frac{{1.00 \times {{10}^{ - 14}}}}{{0.011}} =$ » 12.04

«pH = 14.00 – (–log 0.011)=» 12.04 ✔

Award [3] for correct final answer.

(c)

Sketch the pH curve for the titration of 25.0 cm³ of ethylamine aqueous solution with 50.0 cm³ of butanoic acid aqueous solution of equal concentration. No calculations are required.

[3]

Markscheme

decreasing pH curve ✔

pH close to 7 (6–8) at volume of 25 cm³ butanoic acid ✔

weak acid/base shape with no flat «strong acid/base» parts on the curve ✔

(d)

Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room temperature.

[2]

Markscheme

Any two of:
butanoic acid forms more/stronger hydrogen bonds ✔
butanoic acid forms stronger London/dispersion forces ✔
butanoic acid forms stronger dipole–dipole interaction/force ✔

Accept “butanoic acid forms dimers”

Accept “butanoic acid has larger M_r/hydrocarbon chain/number of electrons” for M2.

Accept “butanoic acid has larger «permanent» dipole/more polar” for M3.

(e.i)

State a suitable reagent for the reduction of butanoic acid.

[1]

Markscheme

lithium aluminium hydride/LiAlH₄ ✔

(e.ii)

Deduce the product of the complete reduction reaction in (e)(i).

[1]

Markscheme