The basicity of  C$_2$H$_2$O$_4$ is

1. Analyze the Question

The question asks for the basicity of the compound $\text{C}_2\text{H}_2\text{O}_4$. Basicity refers to the number of replaceable hydrogen ions ($\text{H}^+$) that one molecule of an acid can release in an aqueous solution.

2. Explain the Concept

• Basicity of an acid: This is determined by the number of ionizable hydrogen atoms present in the acid molecule.
• Structure of $\text{C}_2\text{H}_2\text{O}_4$: This chemical formula represents Oxalic acid. Its structural formula is $(\text{COOH})_2$ or $\text{HOOC-COOH}$.
• In this structure, there are two hydroxyl groups ($-\text{OH}$) attached to carbonyl groups. The hydrogen atoms in these $-\text{OH}$ groups are acidic and can be replaced by a base during a neutralization reaction.

3. Show the Solution

• Step 1: Write out the dissociation equation for Oxalic acid in water: $$\text{C}_2\text{H}_2\text{O}_4(aq) \rightleftharpoons 2\text{H}^+(aq) + \text{C}_2\text{O}_4^{2-}(aq)$$
• Step 2: Count the number of hydrogen ions ($\text{H}^+$) produced. As seen in the equation, one molecule of oxalic acid produces two hydrogen ions.
• Step 3: Since it releases two $\text{H}^+$ ions, the acid is described as dibasic. Therefore, its basicity is 2.

4. Identify the Answer

Comparing our result to the given options: A. 3 B. 4 C. 2 D. 8

The correct option is C.

5. Exam Tip

To quickly determine the basicity of organic acids (carboxylic acids), look at the number of -COOH groups in the formula.

• $\text{CH}_3\text{COOH}$ (Ethanoic acid) has one $-\text{COOH}$ group $\rightarrow$ Basicity = 1 (Monobasic).
• $(\text{COOH})_2$ (Oxalic acid) has two $-\text{COOH}$ groups $\rightarrow$ Basicity = 2 (Dibasic).