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Date November 2018 Marks available 3 Reference code 18N.2.hl.TZ0.6
Level HL Paper 2 Time zone TZ0
Command term Determine Question number 6 Adapted from N/A

Question

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

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

[2]
a.i.

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

[1]
a.ii.

Deduce the average oxidation state of carbon in butanoic acid.

[1]
a.iii.

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]
b.i.

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]
b.ii.

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]
c.

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

[2]
d.

State a suitable reagent for the reduction of butanoic acid.

[1]
e.i.

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

[1]
e.ii.

Markscheme

Butanoic acid:
CH3CH2CH2COOH (aq) + H2O (l) CH3CH2CH2COO (aq) + H3O+ (aq) ✔

 

Ethylamine:
CH3CH2NH2 (aq) + H2O (l) CH3CH2NH3(aq) + OH (aq) ✔

a.i.

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

 

Accept correct diagrams with pi clouds.

a.ii.

–1 ✔

a.iii.

« 1.00 × 10 14 mo l 2 d m 6 0.00192 mol d m 3 » = 5.21 × 10–12 «mol dm–3» ✔

b.i.

«pKb = 3.35, Kb = 10–3.35 = 4.5 × 10–4»

«C2H5NH2 + H2O C2H5NH3+ + OH»

 

Kb =  [ O H ] [ C H 3 C H 2 N H 3  +  ] [ C H 3 C H 2 N H 2 ]

OR

«Kb =» 4.5 × 10–4 = [ O H ] [ C H 3 C H 2 N H 3  +  ] 0.250

OR

«Kb =» 4.5 × 10–4 =  x 2 0.250  ✔


« x = [OH] =» 0.011 «mol dm–3» ✔

 

«pH = –log 1.00 × 10 14 0.011 = » 12.04

OR

«pH = 14.00 – (–log 0.011)=» 12.04 ✔

 

Award [3] for correct final answer.

b.ii.

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 ✔

c.

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 Mr/hydrocarbon chain/number of electrons” for M2.

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

d.

lithium aluminium hydride/LiAlH4

e.i.

butan-1-ol/1-butanol/CH3CH2CH2CH2OH ✔

e.ii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
a.iii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.
[N/A]
e.i.
[N/A]
e.ii.

Syllabus sections

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

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