Date | November 2018 | Marks available | 2 | Reference code | 18N.2.sl.TZ0.6 |
Level | SL | Paper | 2 | Time zone | TZ0 |
Command term | State | 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.
Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room temperature.
State the formula of the salt formed when butanoic acid reacts with ethylamine.
Markscheme
Butanoic acid:
CH3CH2CH2COOH (aq) + H2O (l) CH3CH2CH2COO− (aq) + H3O+ (aq) ✔
Ethylamine:
CH3CH2NH2 (aq) + H2O (l) CH3CH2NH3+ (aq) + OH− (aq) ✔
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.
CH3CH2NH3+ CH3CH2CH2COO−
OR
CH3CH2CH2COO− CH3CH2NH3+
OR
CH3CH2CH2COO− H3N+CH2CH3 ✔
The charges are not necessary for the mark.