Science — CBSE Class 10 board question

(i) Compound 'X' is Ethanoic acid (CH₃COOH).

Structure:
$$\text{CH}_3 - \text{COOH}$$
(Two carbon atoms: a methyl group attached to a carboxyl group –COOH)

(ii) Ethanoic acid is a weak acid — it does not ionise completely in water, so its pH is higher (less acidic) than a mineral acid like HCl or H₂SO₄, which are strong acids and ionise completely, giving a much lower pH.

(iii) The reaction is called Esterification. When ethanoic acid reacts with ethanol (in the presence of conc. H₂SO₄), a sweet-smelling ester is formed:

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

(Ethyl ethanoate is the ester produced.)

(iv) The salt produced is Sodium ethanoate (CH₃COONa). The gas evolved is CO₂.

$$2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

Source: Carbon and its Compounds, Sections 4.3.2–4.3.3

---

• Part (i): Two clues identify X — two carbon atoms + turns blue litmus red (acidic) + used as preservative (vinegar/acetic acid). Students must write both name and structural formula.
• Part (ii): Key distinction: ethanoic acid = weak acid (partial ionisation, higher pH ~3); mineral acids = strong acids (complete ionisation, pH ~1). Do not just say "lower pH" — explain why.
• Part (iii): Name the reaction ("Esterification") and write a balanced equation with catalyst. Forgetting the catalyst (conc. H₂SO₄) or the product name loses marks.
• Part (iv): Examiner looks for the correct salt name and balanced equation. CO₂ turning lime water milky is the identification clue given in the question.