The heat of neutralisation of oxalic acid is `-25.4 kcal mol^(-1)` using strong base, `NaOH`. Hence, the enthalpy change of the process is
`H_(2)C_(2)O_(4) hArr 2H^(o+) +C_(2)O_(4)^(2-)` is

Options

A. (a) `2.0 kcal`
B. (b) `-11.8 kcal`
C. (c) `1.0 kcal`
D. (d) `-1.0 kcal`

Explanation

To solve the problem, we need to determine the enthalpy change (ΔH) for the dissociation of oxalic acid (H₂C₂O₄) into its ions, given the heat of neutralization with NaOH.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The dissociation of oxalic acid can be represented as:
\[
\text{H}_2\text{C}_2\text{O}_4 \rightleftharpoons 2\text{H}^+ + \text{C}_2\text{O}_4^{2-}
...