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

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

Deduce the average oxidation state of carbon in butanoic acid.

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.

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

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.

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

State a suitable reagent for the reduction of butanoic acid.

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

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

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

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

Accept correct diagrams with pi clouds.

–1 ✔

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

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

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

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

OR

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

OR

«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

OR

«pH = 14.00 – (–log 0.011)=» 12.04 ✔

Award [3] for correct final answer.

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 ✔

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.

lithium aluminium hydride/LiAlH₄ ✔

butan-1-ol/1-butanol/CH₃CH₂CH₂CH₂OH ✔