The enthalpy of neutralization of a Weak monoprotic acid (HA) in 1 M solution with a strong base is -55.95 KJ/mol . If the unionized acid required 1.4 KJ/mol heat for it's comptate ionzatation and enthalpy of netralization of the strong monobasic acid with a strong monoacidic base is -57.3 KJ/mol . What is the % ionzation of the weak acid in molar solution ?

Options

A. `1%`
B. `3.57%`
C. `35.7%`
D. 0.1

Explanation

To solve the problem, we need to determine the percentage ionization of a weak monoprotic acid (HA) in a 1 M solution when it reacts with a strong base. We are given the following data:

1. Enthalpy of neutralization of weak acid (HA) with strong base = -55.95 kJ/mol
2. Enthalpy of ionization of the unionized acid (HA) = 1.4 kJ/mol
3. Enthalpy of neutralization of a strong monoprotic acid with a strong base = -57.3 kJ/mol

### Step-by-Step Solution:

**Step 1: Understand the enthalpy changes involved.**

When a weak acid (HA) ionizes, it requires energy (enthalpy of ionization). The reaction can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
The enthalpy change for this ionization is given as \( \Delta H_{ionization} = +1.4 \, \text{kJ/mol} \).

When the weak acid reacts with a strong base (e.g., NaOH), the reaction is:
\[ H^+ + OH^- \rightarrow H_2O \]
The enthalpy change for the neutralization of the strong acid with a strong base is \( -57.3 \, \text{kJ/mol} \).

**Step 2: Set up the equation for the enthalpy of neutralization.**

For the weak acid reacting with a strong base, the enthalpy of neutralization can be expressed as:
\[ \Delta H_{neutralization} = \Delta H_{neutralization \, (strong \, acid)} + \Delta H_{ionization} \times \text{degree of ionization} \]

Let \( x \) be the degree of ionization (as a fraction) of the weak acid. The enthalpy of neutralization for the weak acid can be expressed as:
\[ -55.95 = -57.3 + x \cdot 1.4 \]

**Step 3: Solve for \( x \).**

Rearranging the equation:
\[ -55.95 + 57.3 = x \cdot 1.4 \]
\[ 1.35 = x \cdot 1.4 \]
\[ x = \frac{1.35}{1.4} \]
\[ x = 0.9643 \]

**Step 4: Calculate the percentage ionization.**

The percentage ionization is given by:
\[ \text{Percentage Ionization} = x \times 100 \]
\[ \text{Percentage Ionization} = 0.9643 \times 100 = 96.43\% \]

**Step 5: Determine the final percentage ionization.**

Since the problem asks for the percentage of ionization, we need to find the remaining percentage that is not ionized:
\[ \text{Percentage of non-ionization} = 100 - 96.43 = 3.57\% \]

### Final Answer:
The percentage ionization of the weak acid (HA) in a 1 M solution is **3.57%**.