DP Chemistry (last assessment 2024)

Question 18N.2.hl.TZ0.6b.ii

Date	November 2018	Marks available	[Maximum mark: 3]	Reference code	18N.2.hl.TZ0.6b.ii
Level	hl	Paper	2	Time zone	TZ0
Command term	Determine	Question number	b.ii	Adapted from	N/A

b.ii.

[Maximum mark: 3]

18N.2.hl.TZ0.6b.ii

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

(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.

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Additional higher level (AHL)

Additional higher level (AHL) » Topic 18: Acids and bases » 18.2 Calculations involving acids and bases

Additional higher level (AHL) » Topic 18: Acids and bases

Question 18N.2.hl.TZ0.6b.ii

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