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 analyze the neutralization reaction between methanoic acid (HCOOH) and sodium hydroxide (NaOH). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
When equal volumes of methanoic acid and sodium hydroxide are mixed, a neutralization reaction occurs. The general reaction can be represented as:
\[ \text{HCOOH (aq)} + \text{NaOH (aq)} \rightarrow \text{HCOONa (aq)} + \text{H}_2\text{O (l)} \]
This reaction produces sodium formate (HCOONa) and water.

### Step 2: Identify the Type of Acid and Base
Methanoic acid is a weak acid, while sodium hydroxide is a strong base. This distinction is crucial because it affects the heat evolved during the neutralization process.

### Step 3: 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. For strong acids and strong bases, this value is typically constant and is equal to the heat of formation of water from its ions, denoted as \( x \).

### Step 4: Effect of Weak Acid
Since methanoic acid is a weak acid, it does not completely dissociate in solution. This means that more energy is required to facilitate the neutralization reaction compared to a strong acid. Consequently, the heat evolved during the neutralization of methanoic acid with sodium hydroxide will be less than \( x \).

### Step 5: Conclusion
Therefore, the heat evolved on the neutralization of methanoic acid and sodium hydroxide is less than the heat of formation of water from its ions, which is given as \( x \). Thus, we conclude:
\[ \text{Heat evolved on neutralization} < x \]

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

---