Equal volumes of methanoic acid and sodium hydroxide are mixed. If x is the heat of formation of water from its ions then heat evolved on neutralisation is

Options

A. More than x
B. Equal to x
C. Twice of x
D. Less than x

Answer

To solve the problem, we need to understand the concept of heat of neutralization and how it relates to the heat of formation of water from its ions.

### Step-by-Step Solution:

1. **Identify the Reaction**: When equal volumes of methanoic acid (HCOOH) and sodium hydroxide (NaOH) are mixed, they undergo a neutralization reaction:
\[ \text{HCOOH} + \text{NaOH} \rightarrow \text{HCOONa} + \text{H}_2\text{O} \]

2. **Understand Heat of Neutralization**: The heat of neutralization is defined as the heat evolved when one equivalent of an acid reacts with one equivalent of a base. In this case, we are dealing with a weak acid (methanoic acid) and a strong base (sodium hydroxide).

3. **Consider the Heat of Formation of Water**: Let \( x \) be the heat of formation of water from its ions (H⁺ and OH⁻):
\[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \]
The heat evolved in this reaction is equal to \( x \).

4. **Analyze the Weak Acid Behavior**: Since methanoic acid is a weak acid, it does not completely dissociate in solution. Therefore, additional energy is required to dissociate the weak acid into its ions before neutralization can occur.

5. **Conclusion on Heat Evolved**: Because of the extra energy required for the dissociation of methanoic acid, the total heat evolved during the neutralization will be less than \( x \). Thus, we conclude that the heat evolved on neutralization is:
\[ \text{Heat evolved} < x \]

### Final Answer:
The heat evolved on neutralization is less than \( x \).