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Date November 2018 Marks available 2 Reference code 18N.2.sl.TZ0.6
Level SL Paper 2 Time zone TZ0
Command term Explain 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.

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

[2]
b.

State the formula of the salt formed when butanoic acid reacts with ethylamine.

[1]
c.

Markscheme

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

 

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

a.

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.

b.

CH3CH2NH3+ CH3CH2CH2COO
OR
CH3CH2CH2COO CH3CH2NH3+
OR
CH3CH2CH2COO H3N+CH2CH3

 

The charges are not necessary for the mark.

 

 

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Core » Topic 4: Chemical bonding and structure » 4.4 Intermolecular forces
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