Chemistry 9701 · AS & A Level · Carboxylic acids and derivatives

Carboxylic acids and derivatives — practice question

Ethanoic acid can react with alcohols to make esters, and an equilibrium mixture is produced. This reaction is usually done with an acid catalyst. The equation is $\text{CH}_3\text{CO}_2\text{H} + \text{ROH} \rightleftharpoons \text{CH}_3\text{CO}_2\text{R} + \text{H}_2\text{O}$.
(a)[2]

Write an expression for the equilibrium constant, $K_c$, for this reaction, and state the units clearly.

(b(i))

Calculate the amount, in moles, of $\text{NaOH}$ used in the titration.

(b(ii))

What amount, in moles, of this $\text{NaOH}$ reacted with the hydrogen chloride?

(b(iii))

Write a balanced equation showing the reaction between ethanoic acid and $\text{NaOH}$.

(b(iv))[4]

Hence calculate the amount, in moles, of $\text{NaOH}$ that reacted with the ethanoic acid.

(c(i))

Use your results from (b) to calculate the amount, in moles, of ethanoic acid present at equilibrium. Hence complete the table below.

(c(ii))[3]

Use your results to calculate a value for $K_c$ for this reaction.

(d)[1]

Esters are hydrolysed by sodium hydroxide. During the titration, sodium hydroxide reacts with ethanoic acid and the hydrogen chloride, but not with the ester. Suggest a reason for this.

(e)[2]

What would be the effect, if any, on the amount of ester present if all of the water were removed from the flask and the flask kept for a further week at $25\,^{\circ}\text{C}$? Explain your answer.

Worked solution & mark scheme

This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: So, $K_c = \dfrac{[\mathrm{CH_3CH_2R}][\mathrm{H_2O}]}{[\mathrm{CH_3CH_2H}][\mathrm{ROH}]}$

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